WPCL 2BJ|x! x  Њ`SOURCES page ! x  tЊ  t 7.ARITHMETIC & NUMBER-THEORETIC RECREATIONS 7.A.FIBONACCI NUMBERS  tWe use the standard form: F0 = 0, F1 = 1, Fn+1 = Fn + Fn1, with the auxiliary Lucas numbers being given by: L0 = 2, L1 = 1, Ln+1 = Ln + Ln1.  t Parmanand Singh. The socalled Fibonacci numbers in ancient and medieval India. HM 12 (1985) 229244. In early Indian poetry, letters had weights of 1 or 2 and meters were classified both by the number of letters and by the weight. Classifying by weight gives the number of sequences of 1s and 2s which add to the weight n and this is Fn+1. Pingala (c450) studied prosody and gives cryptic rules which have been interpreted as methods for generating the next set of sequences, either classified by number of letters or by weight and several later writers have given similar rules. The generation implies Fn+1 = Fn + Fn1. Virah]nka (c7C) is slightly more explicit. Gop]la (c1134) gives a commentary on Virah]nka which explicitly gives the numbers as 3, 5, 8, 13, 21. Hemacandra (c1150) states "Sum of the last and last but one numbers ... is ... next." This is repeated by later authors. The Pr]krta Paingala (c1315) gives rules for finding the kth sequence of weight n and for finding the position of a particular sequence in the list of sequences of weight n and the positions of those sequences having a given number of 2s (and hence a given number of letters). It also gives the relation Fn+1 = $i BC(ni,i). Narayana Pandita (= N]r]yana Pandita) 's Ganita Kaumudi (1356) studies additive sequences in chap.13, where each term is the sum of the last q terms. He gives rules which are equivalent to finding the coefficients of (1 + x + ... + xq1)p and relates to ordered partitions using 1, 2, ..., q. Narayana Pandita (= N]r]yana Pandita). Ganita Kaumudi (1356). Part I, (p. 126 of the Sanskrit ed. by P. Dvivedi, Indian Press, Benares, 1942), ??NYS quoted by Kripa Shankar Shukla in the Introduction to his edition of: Narayana Pandita (= N]r]yana Pandita); Bjaganit]vatamsa; Part I; Akhila Bharatiya Sanskrit Parishad, Lucknow, 1970, p. iv. "A cow gives birth to a calf every year. The calves become young and themselves begin giving birth to calves when they are three years old. Tell me, O learned man, the number of progeny produced during twenty years by one cow." WESTERN HISTORIES H. S. M. Coxeter. The golden section, phyllotaxis, and Wythoff's game. SM 19 (1953) 135-143. Sketches history and interconnections. H. S. M. Coxeter. Introduction to Geometry. Wiley, 1961. Chap. 11: The golden section and phyllotaxis, pp. 160172. Extends his 1953 material. Maxey Brooke. Fibonacci numbers: Their history through 1900. Fibonacci Quarterly 2:2 (1964) 149-153. Brief sketch, with lots of typographical errors. Doesn't know of Bernoulli's work. Leonard Curchin & Roger HerzFischler. De quand date le premier rapprochement entre la suite de Fibonacci et la division en extr+me et moyenne raison? Centaurus 28 (1985) 129138. Discusses the history of the result that the ratio Fn+1/Fn approaches C. Pacioli and Kepler, described below, seem to be the first to find this. Roger Herz-Fischler. Letter to the Editor. Fibonacci Quarterly 24:4 (1986) 382. Roger HerzFischler. A Mathematical History of Division in Extreme and Mean Ratio. Wilfrid Laurier University Press, Waterloo, Ontario, 1987. Retitled: A Mathematical History of the Golden Number, with new preface and corrections and additions, Dover, 1998. Pp. 157162 discuss early work relating the Fibonacci sequence to division in extreme and mean ratio. 15 pages of references. Georg Markovsky. Misconceptions about the Golden Ratio. CMJ 23 (1992) 219. This surveys many of the common misconceptions e.g. that C appears in the Great Pyramid, the Parthenon, Renaissance paintings and/or the human body and that the Golden Rectangle is the most pleasing with 59 references. He also discusses the origin of the term 'golden section', sketching the results given in HerzFischler's book. Thomas Koshy. Fibonacci and Lucas Numbers with Applications. WileyInterscience, Wiley, 2001. Claims to be 'the first attempt to compile a definitive history and authoritative analysis' of the Fibonacci numbers, but the history is generally secondhand and marred with a substantial number of errors, The mathematical work is extensive, covering many topics not organised before, and is better done, but there are more errors than one would like. Ron Knott has a huge website on Fibonacci numbers and their applications, with material on many related topics, e.g. continued fractions, !, etc. with some history. www.ee.surrey.ac.uk/personal/r.knott/fibonacci/fibnat.html . Fibonacci. 1202. Pp. 283-284 (S: 404405): Quot paria coniculorum in uno anno ex uno pario germinentur [How many pairs of rabbits are created by one pair in one year]. Rabbit problem the pair propagate in the first month so there are Fn+2 pairs at the end of the nth month. (English translation in: Struik, Source Book, pp.2-3.) I have colour slides of this from L.IV.20 & 21 and Conventi Soppresi, C. I. 2616. This is on ff.130r130v of L.IV.20, f. 225v of L.IV.21, f. 124r of CS.C.I.2616. Unknown early 16C annotator. Marginal note to II.11 in Luca Pacioli's copy of his 1509 edition of Euclid. Reproduced and discussed in Curchin & HerzFischler and discussed in HerzFischler's book, pp. 157158. II.11 involves division in mean and extreme ratio. Uses 89, 144, 233 and that 1442 = 89 * 233 + 1. Also refers to 5, 8, 13. Gori. Libro di arimetricha. 1571. F. 73r (p.81). Rabbit problem as in Fibonacci. J. Kepler. Letter of Oct 1597 to Mstlin. ??NYS described in HerzFischler's book, p. 158. This gives a construction for division in extreme and mean ration. On the original, Mstlin has added his numerical calculations, getting 1/C = .6180340, which HerzFischler believes to be the first time anyone actually calculated this number. J. Kepler. Letter of 12 May 1608 to Joachim Tanckius. ??NYS described in HerzFischler (1986), Curchin & HerzFischler and HerzFishler's book, pp. 160161. Shows that he knows that the ratio Fn+1/Fn approaches C and that Fn2 + (1)n = Fn1Fn+1. J. Kepler. The Six-Cornered Snowflake. Op. cit. in 6.AT.3. 1611. P.12 (20-21). Mentions golden section in polyhedra and that the ratio Fn+1/Fn approaches C. See HerzFischler's book, p. 161. Albert Girard, ed. Les uvres Mathematiques de Simon Stevin de Bruges. Elsevier, Leiden, 1634. Pp. 169-170, at the end of Stevin's edition of Diophantos (but I have seen other page references). Notes the recurrence property of the Fibonacci numbers, starting with 0, and asserts that the ratio Fn+1/Fn approaches the ratio of segments of a line cut in mean and extreme ratio, i.e. C, though he doesn't even give its value but he says 13,13, 21 'rather precisely constitutes an isosceles triangle having the angle of a pentagon'. HerzFischler's book, p. 162, notes that Girard describes it as a new result and includes 0 as the starting point of the sequence. Abraham de Moivre. The Doctrine of Chances: or, A Method of Calculating the Probability of Events in Play. W. Pearson for the Author, London, 1718. Lemmas II & III, pp.128-134. Describes how to find the generating function of a recurrence. One of his illustrations is the Lucas numbers for which he gets: x + 3x2 + 4x3 + 7x4 + 11x5 + ... = (2x + x2)/(1 x x2). However, he does not have the Fibonacci numbers and he does not use the generating function to determine the individual coefficients of the sequence. In Lemma III, he describes how to find the recurrence of p(n) an where p(n) is a polynomial. Koshy [p. 215] says De Moivre invented generating functions to solve the Fibonacci recurrence, which seems to be reading much more into De Moivre than De Moivre wrote. The second edition is considerably revised, cf below. Daniel Bernoulli. Observationes de seriebus quae formantur ex additione vel substractione quacunque terminorum se mutuo consequentium, ubi praesertim earundem insignis usus pro inveniendis radicum omnium aequationum algebraicarum ostenditur. Comm. Acad. Sci. Petropolitanae 3 (1728(1732)) 85-100, ??NYS. = Die Werke von Daniel Bernoulli, ed. by L. P. Bouckaert & B. L. van der Waerden, Birkhuser, 1982, vol. 2, pp. 49+??. Section 6, p. 52, gives the general solution of a linear recurrence when the roots of the auxiliary equation are distinct. Section 7, pp. 52-53, gives the 'Binet' formula for Fn. [Binet's presentation is so much less clear that I suggest the formula should be called the Bernoulli formula.] Abraham de Moivre. The Doctrine of Chances: or, A Method of Calculating the Probability of Events in Play. 2nd ed, H. Woodfall for the Author, London, 1738. Of the Summation of recurring Series, pp. 193206. This is a much revised and extended version of the material, but he says it is just a summary, without demonstrations, as he has given the demonstrations in his Miscellanea Analytica of 1730 (??NYS). Gives the generating functions for various recurrences and even for a finite number of terms. Prop. VI is: In a recurring series, any term may be obtained whose place is assigned. He assumes the roots of the auxiliary equation are real and distinct. E.g., for a second order recurrence with distinct roots m, p, he says the general term is Amn + Bpn where he has given A and B in terms of the first two values of the recurrence. He even gives the general solution for a fourth order recurrence and expresses A, B, C, D in terms of the first four values of the recurrence. Describes how to take the even terms and the odd terms of a recurrence separately and how to deal with sum and product of recurrences. R. Simson. An explication of an obscure passage in Albert Girard's commentary upon Simon Stevin's works. Phil. Trans. Roy. Soc. 48 (1753) 368-377. Proves that Fn2Ġ+(-1)nĠ=Fn1Fn+1. This says that the triple Fn1, Fn, Fn+1 "nearly express the segments of a line cut in extreme and mean proportion, and the whole line;" from which he concludes that the ratio Fn+1/Fn does converge to C. HerzFischler's book, p. 162, notes that his proof is essentially an induction. (He also spells the author Simpson, but it is definitely Simson on the paper.) [Koshy, p. 74, says Fn2Ġ+(-1)nĠ=Fn1Fn+1 was discovered in 1680 by Giovanni Domenico Cassini, but he gives no reference and neither Poggendorff nor BDM help to determine what paper this might be.] Ch. Bonnet. Recherches sur l'usage des feuilles dans les plantes. 1754, pp. 164-188. Supposed to be about phyllotaxis but only shows some spirals without any numbers. Refers to Calandrini. Nice plates. Master J. Paty (at the Mathematical Academy, Bristol), proposer; W. Spicer, solver. Ladies' Diary, 176869 = T. Leybourn, II: 293, quest. 584. Cow calves at age two and every year thereafter. How many offspring in 40 years? Answer is: 0 + 1 + 1 + 2 + 3 + ... + F39. We would give this as: F41 1, but he gets it as: 2F39 + F38 1. Eadon. Repository. 1794. P. 389, no. 47. Same as the previous problem. Jackson. Rational Amusement. 1821. Curious Arithmetical Questions, no. 32, pp. 22 & 82. Similar to Fibonacci, with a cow and going for 20 generations. Martin Ohm. Die reine ElementarMathematik. 2nd ed., Jonas VerlagsBuchhandlung, Berlin, 1835. P. 194, footnote to Prop. 5. ??NYS extensively discussed in HerzFischler's book, p. 168. This is the oldest known usage of 'goldene Schnitt'. It does not appear in the 1st ed. of 1826 and here occurs as: "... nennt man wohl auch den goldenen Schnitt" (... one also appropriately calls [this] the golden section). The word 'wohl' has many, rather vague, meanings, giving different senses to Ohm's phrase. HerzFischler interprets it as 'habitually', which would tend to imply that Ohm and/or his colleagues had been using the term for some time. I don't really see this meaning and interpreting 'wohl' as 'appropriately' would give no necessity for anyone else to know of the phrase before Ohm. However the term is used in several other German books by 1847. [Incidentally, this is not the Ohm of Ohm's Law, but his brother.] A. F. W. Schimper & A. Braun. Flora. 1835. Pp. 145 & 737. ??NYS J. Binet. M)moire sur l'integration des )quations lin)aires aux diff)rences finies, d'un ordre quelconque, ! coefficients variables. (Extrait par l'auteur). CR Acad. Sci. Paris 17 (1843) 559-567. States the Binet formula as an example of a general technique for solving recurrences of the form: v(n+2) = v(n+1) + r(n)v(n), but the general technique is not clearly described, nor is the linear case. B. Peirce. Mathematical investigation of the fractions which occur in phyllotaxis. Proc. Amer. Assoc. Adv. Sci. 2 (1849) 444-447. Not very interesting. Gustav Theodor Fechner. Vorschule der sthetik. Breitkopf & Hrtel, Leipzig, 1876. ??NYS. Origin of the aesthetic experiments on golden rectangles. Koshy [p. 5] says Lucas originally called Fn the 's)rie de Lam)', but introduced the name Fibonacci numbers in May 1876. However, he doesn't give a reference. There are several papers by Lucas which might be the desired paper. Note sur le triangle arithm)tique de Pascal et sur la s)rie de Lam). Nouvelle Correspondence Math)matique 2 (1876) 7075; which might be the desired paper. L'arithm)tique, la s)rie de Lam), le probl/me de BehaEddin, etc. Nouvelles Annales de Math)matiques 15 (1876) 20pp. (douard Lucas. Th)orie des fonctions num)riques simplement p)riodiques. AJM 1 (1878) 184240 (Sections 123) & 289321 (Sections 2430). [There is a translation by Sidney Kravitz of the first part as: The Theory of Simply Periodic Numerical Functions, edited by Douglas Lind, The Fibonacci Association, 1969. Dickson I 400, says this consists of 7 previous papers in Nouv. Corresp. Math. in 18771878 with some corrections and additions. Robert D. Carmichael; Annals of Math. (2) 15 (1913) 3070, ??NX, gives corrections.] The classic work which begins the modern study of recurrences. Koshy, p. 273, says Adolf Zeising's Der goldene Schnitt of 1884 put forth the theory that "the golden ratio is the most artistically pleasing of all proportions ...." But cf Fechner, 1876. Pearson. 1907. Part II, no. 63: A prolific cow, pp. 126 & 203. Same as Fibonacci's rabbits, but wants the total after 16 generations. Koshy, p. 242, asserts that Mark Barr, an American mathematician, introduced the symbol C (from Phidias) for the Golden Ratio, (1 + 5)/2, about 1900, but he gives no reference. Coxeter, 1953, takes ) from the initial letter of ), the Greek word for section, but I have no idea if this was used before him. There is a magic trick where you ask someone to pick two numbers and extend them to a sequence of ten by adding the last two numbers each time. You then ask him to add up the ten numbers and you tell him the answer, which is 11 times the seventh number. In general, if the two starting numbers are A and B, the nth term is Fn2A + Fn1B and the sum of the first 2n terms is F2nA + (F2n+1ĩ1)B = Ln (FnA + Fn+1B), but only the case n = 5 is interesting! I saw Johnny Ball do this in 1989 and I have found it in: Shari Lewis; Abracadabra! Magic and Other Tricks; (World Almanac Publications, NY, 1984); Puffin, 1985; Sum trick!, p. 14, but it seems likely to be much older. 7.B.JOSEPHUS OR SURVIVOR PROBLEM  tSee Tropfke 652. This is the problem of counting out every k-th from a circle of n. Early versions counted out half the group; later authors and the Japanese are interested in the last man the survivor. Euler (1775) seems to be the first to ask for the last man in general which we denote as L(n, k). Cardan, 1539, is the first to associate this process with Josephus. Some later authors derive this from the Roman practice of decimation. For last man versions, see the general entries and: Michinori?, Kenk, Cardan, Coburg, Bachet, van Etten, Yoshida, Muramatsu, Schnippel, Ozanam (1696 & 1725), LesAmusemens, Fujita, Euler, Miyake, Matuoka, Boy's Own Book, Nuts to Crack, TheSociable, Indoor & Outdoor, Secret Out (UK), Leske, Le Vallois, Hanky Panky, Kemp, Mittenzwey, Gaidoz, Ducret, Lemoine, Akar et al, Lucas, Schubert, Busche, Tait, Ahrens, Rudin, MacFhraing, Mendelsohn, Barnard, Zabell, Richards, Dean, Richards,  t 2 to last, counted by 9s: Boy's Own Book, 3 to last, counted by 9s: Boy's Own Book, 4 to last, counted by 9s: Boy's Own Book, 5 to last, counted by 9s: Boy's Own Book, 6 to last, counted by 9s: Boy's Own Book, 7 to last, counted by 9s: Boy's Own Book, 9 to last, counted by 9s: Boy's Own Book, 10 to last, counted by 9s: Boy's Own Book, 11 to last, counted by 9s: Boy's Own Book, 12 to last, counted by 9s: Boy's Own Book, Secret Out (UK), 12 to last, counting number unspecified: Coburg, 13 to last, counted by 2s: Ducret, Leeming, 13 to last, counted by 9s: Boy's Own Book, Secret Out (UK), Leske, Rudin, 14 to all!, counted by 6s: Secret Out, 14 to last, counted by 10s: Mittenzwey, 17 to last, counted by 3s: Barnard, 21 to last, counted by 5s: Hyde, 21 to last, counted by 7s: Nuts to Crack, The Sociable, Indoor & Outdoor, Hanky Panky, H.D. Northrop, 21 to last, counted by 8s: Mittenzwey, 21 to last, counted by 10s: Hyde, 24 to last, counted by 9s: Kemp, 28 to last, counted by 9s: Kemp, 30 to last, counted by 9s: Schnippel, 30 to last, counted by 10s: see entries in next table for 15 & 15 counted by 10s 40 to last man, counted by 3s: van Etten (erroneous), 41 to last man, counted by 3s: van Etten, Ozanam (1725), Vinot, Ducret, Lucas (1895), General case: Euler, Lemoine, Akar et al., Schubert, Busche, Tait, Ahrens, MacFhraing, Mendelsohn, Robinson, Jak;bczyk, Herstein & Kaplansky, Zabell, Richards,  t There are a few examples where one counts down to the last two persons see references to Josephus and: Pacioli, Muramatsu, Mittenzwey, Ducret, Les Bourgeois Punis. Almost all the authors cited consider 15 & 15 counted by 9s, so I will only index other versions.  t 2 & 2 counted by 3s: Ball (1911), 2 & 2 counted by 4s: Ball (1911), 3 & 3 counted by 7s: Ball (1911), 3 & 3 counted by 8s: Ball (1911), 4 & 4 counted by 2s: Leeming, 4 & 4 counted by 5s: Ball (1911), 4 & 4 counted by 9s: Ball (1911), 5 & 5 counted by ??: Dudeney (1905), Pearson, Ball (1911), Ball (1920), Shaw, 6 & 6 counted by ??: Dudeney (1900), 8 & 2 counted by ??: Les Bourgeois Punis, 8 & 8 counted by 8s: Kanchusen, 8 & 8 counted by ??: Dudeney (1899), 12 & 12 counted by 6s: Harrison, 15 & 15 counted by 3s: Tartaglia, Alberti, 15 & 15 counted by 4s: Tartaglia, 15 & 15 counted by 5s: Tartaglia, 15 & 15 counted by 6s: AR, Codex lat. Monacensis 14908, Tartaglia, 15 & 15 counted by 7s: Tartaglia, Schnippel, 15 & 15 counted by 8s: Codex lat. Monacensis 14836, AR, Codex lat. Monacensis 14908, Tartaglia, Alberti, 15 & 15 counted by 10s: Michinori?, Reimar von Zweiter, AR, Codexlat.Monacensis14908, Chuquet, Tartaglia, Buteo, Hunt, Yoshida, Muramatsu, Wingate/Kersey, Schnippel, Alberti, Shinpen Kinkoki, Fujita, Miyake, Matuoka, SanpoChieBukuro, Hoffmann, Brandreth, Benson, Williams, Collins, Dean. (Almost all of these actually continue to the last person.) 15 & 15 counted by 11s: Tartaglia, Schnippel, 15 & 15 counted by 12s: AR, Codex lat. Monacensis 14908, Tartaglia, 15 & 15 counted by other values, not specified ??check: Codex lat. Monacensis 14836, Meermanische Codex, at-Tilimsni, Bartoli, Murray 643, Chuquet, Keasby, 17 & 15 counted by 10s: Schnippel, 17 & 15 counted by 12s: Mittenzwey, 18 & 2 counted by 12s: Pacioli, Rudin, 18 & 6 counted by 8s: Manuel des Sorciers, 18 & 18 counted by 9s: Chuquet, 24 & 24 counted by 9s: Chuquet, 30 & 2 counted by 7s: Pacioli, 30 & 2 counted by 9s: Pacioli, 30 & 6 counted by 10s: Ducret, 30 & 10 counted by 12s: Endless Amusement II, Magician's Own Book, The Sociable, Boy's Own Conjuring Book, Lucas (1895), 30 & 30 counted by 12s: Sarma, 36 & 4 counted by 10s: Jackson, n2ĩn+1 & n1 counted by n: Lucas (1894), CesarA, Franel, Akar,  t Many authors provide a mnemonic for the case of 15 and 15 counted by 9s. In this case, the longest group of the same type is five, so a common device is to encode the numbers 1, 2, 3, 4, 5 by the vowels a,e, i, o, u and then produce a phrase with the vowels in the correct order. I will call this a vowel mnemonic. The most popular form is: Populeam virgam Mater Regina ferebat, giving the numerical sequence: 4,5, 2, 1, 3, 1, 1, 2, 2, 3, 1, 2, 2, 1. The first group of 4 are good guys, followed by 5 bad guys, etc. Below I list the mnemonics and where they occur, but I did not always record them in my notes below, so I must check a number of the sources again ?? the classification was inspired by seeing that Franci (op. cit. in 3.A) describes a vowel mnemonic in Benedetto da Firenze which I had overlooked. Ahrens gives many more verse and vowel mnemonics to be added below. Hyde gives an Arabic mnemonic due to alSafadi using the first letters of the Arabic alphabet: a, b, gj, d, h.  tDahbagja Ababgja Baba [= Da h b a gj a A ba b gj a Ba b a]: Hyde from alSafadi. Unspecified(?) verse mnemonic: ibn Ezra Populea irgam mater regina reserra: Pacioli; Populea virga pacem regina ferebat: Mingu)t Populeam jirgam mater Regina ferebat: Badcock Populeam virgam mater Regina ferebat: van Etten; Hunt; Schnippel; Ozanam1725; LesAmusemens; Hooper; Jackson; Manuel des Sorciers; Boy's Own Book; TheSociable; Le Vallois; Gaidoz; Lucas Populeam virgam mater regina reserrat: Agostini's version of Pacioli; Populeam virgam Mater Regina tenebat: Hyde from Wit's Interpreter; Murphy; Schnippel/Bolte Mort tu ne failliras pas en me liurant le trespas: van Etten Mort, tu ne falliras pas En me livrant au tr)pas: Manuel des Sorciers; Mort, tu ne falliras pas. En me livrant le tr)pas: Schnippel/Bolte; Ozanam 1725; LesAmusemens; The Sociable; Le Vallois (without the first comma); Ducret; Lucas On tu ne dai la pace ei la rendea: Schnippel/Bolte Gott schuf den Mann in Amalek, der (or den) Israel bezwang: Schnippel Gott schlug den Mann in Amalek, den Israel bezwang: Schnippel/Bolte So du etwan bist gfalln hart, Stehe widr, Gnade erwart: Schnippel/Bolte Non dum pena minas a te declina degeas: Schnippel/Bolte Nove la pinta d! e certi mantena: Benedetto da Firenze From member's aid and art, Never will fame depart: Schnippel/Bolte From numbers, aid and art / Never will fame depart: Wingate/Kersey From numbers, aid, and art, Never will fame depart: Ingleby; Jackson; Rational Recreations From number's aid and art, Never will fame depart: Gaidoz From numbers aid and art / Never will fame depart: The Sociable I have only one example of a mnemonic for 15 & 15 counted by 10s. Rex Paphicum Gente Bonadat Signa Serena: Hunt See 5.AD for the general problem of stacking a deck to produce a desired effect. Josephus. De Bello Judaico. c80. Book III, chap. 8, sect. 7. (Translated by Whiston or by Thackeray (Loeb Classical Library, Heinemann, London, 1927, vol. 2, pp. 685-687.)) (Many later authors cite Hegesippus which is a later version of Josephus.) This says that Josephus happened to survive "by chance or God's providence". H. St. J. Thackeray. Josephus, the Man and the Historian. Jewish Institute Press, NY, 1929, p.14. Comments on the Slavonic text, which says that Josephus "counted the numbers with cunning and thereby misled them all" but gives no indication how. Ahrens. MUS II. 1918. Chap. XV: Das Josephsspiel, pp. 118169. This is the most extended and thorough discussion of this problem and its history. I have used it as the basis of this section. He gives a rather complex method, based on work of Busche, Schubert and Tait, for determining the last man, or any other man in the sequence of counting out, which I never worked through, but which is clearly explained under Richards (1999/9). Gerard Murphy. The puzzle of the thirty counters. B)aloideas The Journal of the Folklore of Ireland Society XII (1942) 3-28. In this work and the material cited (mostly ??NYS), the problem of 15 and 15 counted by 9s is shown to have the medieval name Ludus Sancti Petri = St. Peters-Spiel = St. Peter's Lake (lake being an Old English word for game [or AngloSaxon for 'to play']) = Sankt Peter Lek or Sankt Pders Lek (in Swedish). Murphy cites Schnippel & Bolte to assert that it was known to the Arabs in the 14C cf below. He says the usual European version has Christians and Jews on a ship, with St. Peter present and suggesting the counting out process. [I had forgotten that such versions occur in Ahrens, MUS II 130.] However, Murphy was unable to consult MUS, so his background is not as complete as it might be. Murphy demonstrates that the problem was recently well-known in both Scots and Irish Gaelic in a form where a woman has to choose between two groups of warriors seated in a circle, with emotional reasons for her preference. The solution is given in a vernacular mnemonic, using actual numbers as in early Latin forms, while later Latin and vernacular forms used vowel mnemonics. He gives an Irish reconstruction, with English translation, based on several 18C MSS whose texts he estimates as 13C to 17C, probably 16C. This is titled: Goid Fhinn Agus Dubhin Anso (Here is the Thieving of Fionn and Dubhn). One MS has the Latin subtitle: Populeam virgam Mater Regina tenebat, which is a common Latin vowel mnemonic. One of Murphy's sources says this refers to the Queenly Mary appearing to the ship's captain and holding a poplar rod. Murphy also gives an extended Irish story (3pp) built around the problem: Ceann Dubhrann na Ndumhchann Bn (Ceann Dubhrann of the White Sandhills). The Gaelic names Fionn and Dubhn are derived from 'fionn' and 'dub' meaning 'white' and 'black'. Murphy gives a contemporary Irish version on board a ship with a white captain and a black wife and a crew of 15 and 15, with half having to go overboard due to lack of food. He sketches numerous other Irish and Scots version with varying combinations of details, but using essentially the same verse mnemonic. Murphy cites a study by Manitius of a 9C MS (Bib. Nat. Paris, No. 13029) where the problem begins "Quadam nocte niger dub nomine, candiduus alter" (One night a black man named Dub and another [named] White). They have to choose between the blacks and the whites to keep watch. Cf. Codex Einsidelensis No. 326 below. The MS ends with the prose line "These two Irish soldiers, one named 'Find' the other 'Dub', were engaged in hunting. 'Find' means "white", 'dub' "black"." The 12C Rouen MS No.1409 attributes the problem to a Clemens Scottus, which Murphy interprets as Clement the Irishman. The 12C MS Bib. Nat. Paris No. 8091 attributes it to a Thomas Scottus. Murphy concludes that the problem has an Irish origin, c800. He gives what he believes to be the earliest Latin form, basically Bib. Nat. Paris No. 13029, and opines there must have been an Irish predecessor. Codex Einsidelensis No. 326. 10C. F. 88'. Latin verse. Published by Th. Mommsen, Handschriftliches. Zur lateinischen Anthologie. Rheinischen Museum fGr Philologie (NS) 9 (1854) 296-301, with material of interest on pp. 298299. Latin given in: M.Curtze, Bibliotheca Math. (2) 9 (1895) 34-35. Latin & German in MUS II 123-125. Begins: "Quadam nocte niger dux nomine, candidus alter". 15 white & 15 black soldiers, half to keep watch, counted off by 9. The colours refer to clothing, not skin! Codex lat. Monacensis 14836. 11C. F. 80' gives rules for 15 and 15 counted by 9 (though this value is not specified) and mentions counting by 8 and other values. No mention of what is being counted. Quoted and discussed by: M. Curtze; Zur Geschichte der Josephspiels; Bibliotheca Math. (2) 8 (1894) 116 and in: Die Handschrift No. 14836 der K?nigl. Hof- und Staats-bibliothek zu MGnchen; AGM 7 (1895) 105 & 111-112 (Supplement to Zeitsch. fGr Math. und Physik 40 (1895)). Codex Bernensis 704. 12C. Published by: Hermann Hage; Carmina medii aevi maximam partem inedita; Ex Bibliothecis Helveticus collecta; Bern, 1877; no. 85, pp. 145-146. ??NYS. Latin in: Curtze, op.cit. at Codex Einsidelensis, pp. 35-36; and in: MUS II 127. Jews & Christians. Ahrens, MUS II 118-147, gives many further references from 10-13C. Originals ??NYS.  x  thMeermanische Codex, 10C. Mentions counting by other values. Leiden Miscellancodex, 12C Basel Miscellancodex, 13C  x  tMichinori Fujiwara (11061159). This work is lost, but has been conjectured to contain a form of the problem see under Kenk, c1331, and Yoshida, 1634. Rabbi Abraham ben Ezra. Ta'hbula (or TachbEla), c1150. ??NYS described in: Moritz Steinschneider; Abraham ibn Ezra (Abraham Judaeus, Avenare); Zur Geschichte der mathematischen Wissenschaft im XII Jahrhundert; Zeitschr. fGr Math. und Physik 25 (1880): Supp: AGM 3, Part II (1880). The material is Art. 20, pp. 123-124. 15 students and 15 goodfornothings on a ship, counted by 9s. This seems to be the first extant example on board a ship. Verse mnemonic, which Steinschneider says is not original. Steinschneider cites further sources. Smith & Mikami, p. 84, say ben Ezra died in 1067 ?? Reinmar von Zweter. Meisterlied: "Ander driu, wie man juden und cristen Ez zelt". 13C. (In MUS II 128.) Jews and Christians on a ship, counts by 10. Kenk, also known as Kenk Yoshida or Urabe no Kaneyoshi or Yoshida no Kaneyoshi (12831350). Tsurezuregusa. c1331. Translated by Donald Keene as: Essays in Idleness The Tsurezuregusa of Kenk; Columbia Univ. Press, NY, 1967. (Kenk was the author's monastic name. His lay name was Urabe no Kaneyoshi. He lived for a long time at Yoshida in Kyoto.) This book is one of the classics of Japanese literature, consisting of 243 essays, ranging from single sentences to several pages. The most common themes of these relate to the impermanence of life and the vanity of man. In Japanese, the Josephus problem is called Mamakodate or Mamakodate San (or Mamako tate no koto cf Matoka, 1808) or Mamagodate (Scheme to benefit the stepchildren or Stepchild disposition). It is said to have been in the lost work of Michinori Fujiwara (1106-1159), qv. The word Mamagodate first occurs in essay 137 of Kenk, pp. 115121 in Keene's version (including a doublepage illustration which doesn't depict the problem), whose beginning is characteristic of Kenk's style: "Are we to look at cherry blossoms only in full bloom, the moon only when it is cloudless? To long for the moon while looking on the rain, to lower the blinds and be unaware of the passing of the spring these are even more deeply moving." The passage of interest is toward the end, on p. 120 of Keene: "When you make a mamagodate1 with backgammon counters, at first you cannot tell which of the stones arranged before you will be taken away. Your count then falls on a certain stone and you remove it. The others seem to have escaped, but as you renew the count you will thin out the pieces one by one, until none is left. Death is like that." The footnote refers to counting 15 and 15 by 10s, so that 14 white stones are eliminated, then the counting is reversed and all the black stones are eliminated. "The Japanese name mamagodate (stepchild disposition) derives from the story of a man with fifteen children by one wife and fifteen by another; his estate was disposed of by means of the game, one stepchild in the end inheriting all." Kenk's text clearly shows he was familiar with the process of counting to the last man and the use of the name indicates that he was familiar with the version mentioned in the footnote, though its earliest explicit appearance in Japan is in Yoshida, 1634, qv. My thanks to Takao Hayashi for the reference to Keene. Thomas Hyde. Mandragorias seu Historia Shahiludii, .... (= Vol. 1 of De Ludis Orientalibus, see 4.B.5 for vol. 2.) From the Sheldonian Theatre (i.e. OUP), Oxford, 1694. Prolegomena curiosa. The initial material and the Prolegomena are unpaged but the folios of the Prolegomena are marked (a), (a 1), .... The material is on (e 1).v (e 2).v, which are pages 3436 if one starts counting from the beginning of the Prolegomena. Cited by Bland (loc. cit. in 5.F.1 under Persian MS 211, p. 31); Ahrens (MUS II 136) & Murray 280. Several citations are to ii.23, which may be to the 1767 reprint of Hyde's works. Hyde asserts that the problem of the ship with 15 Moslems and 15 Christians on a ship, counted by 9s, was given by alSafadi (Salhadd3n as-Safad3 = alSphadi =AlSphadi) (d. 1363) in his Lmiyato l Agjam (variously printed in the text). This must be his Sharh Lm3yat al-Ajam of c1350. Hyde gives an Arabic mnemonic using the first five letters of the Arabic alphabet, which he transliterates as: Dahbagja Ababgja Baba [= Da h b a gj a A ba b gj a Ba b a]. He says the problem occurs in an English book called Wit's Interpreter (??NYS) (8oS.87.Art) where the mnemonic Populeam virgam mater Regina tenebat is given. He then says that the problem is also described in 'Megjdium & Abulphedam' p. 43 of his main text identifies Abulpheda as a prince born in 672 AH ?? Shihbadd3n AbEl-Abbs Ahmad ibn Yahya ibn Ab3 Hajala at-Tilimsni alH-anbal3 (??sp). Kitb anmEdhaj al-qitl fi lab ash-shatranj [Book of the examples of warfare in the game of chess]. c1370. Copied by Muhammed ibn Ali ibn Muhammed al-Arzag3 in 1446. This is the second of Dr. Lee's MSS, described in 5.F.1, denoted Man. by Murray. Murray 280 says "Man. 36-45 relate to as-Safad3's problem of the ship (see Hyde, ii.23)", described by Murray as 15 Christians and 15 Muslims counted by n. Bland has "the wellknown problem of the Ship, first as described by Safadi, and then in other varieties. (Hyde, p. 23.)" Murray 620 says the problem is of Muslim origin and says it appears in the c1530 Italian version of the Bonus Socius collection which Murray denotes It. (See 5.F.1 under Bonus Socius.) Murray refers to 15 & 15 counted by 9s, but it is not clear if this refers to this particular MS. Murray 622 cites MS Sloan 3281 in the BM, 14C, as giving the Latin mnemonic solution. Bartoli. Memoriale. c1420. F. 100r (= Sesiano p. 135). Il giuocho de' Cristiani contra Saracin. 15 Christians and 15 Saracens the text ends in the middle of the statement of the problem. AR. c1450. Prob. 80, pp. 52, 181-182 & 229. 15 Christians and 15 Jews. Gives only mnemonics for counting by 10, 9, 8, 6 or 12. Codex lat. Monacensis 14908. c1460. F. 76 gives mnemonics for 15 Jews and 15 Christians counted by 6, 8, 9, 10, 12. Quoted and discussed by Curtze, opp. cit. under Cod. lat. Mon. 14836, above. [In the first paper, the codex number is misprinted as 14809.] Benedetto da Firenze. c1465. Pp. 142-143. 15 Christians and 15 Jews on a boat counted by 9s. Vowel mnemonic: Nove la pinta d! e certi mantena. Diagrammatic picture on p.143. Murray 643 says the MS Lasa version, c1475, of the Civis Bononiae collection (described in 5.F.1 under Civis Bononiae) has "16 diagrams of the 'ship' puzzle under different conditions". Chuquet. 1484. Prob. 146. English in FHM 228230, with reproduction of the original on p. 229. 15 Jews and 15 Christians on a ship, counting by 9s. Says one can have 18 or 24 of each and can count by 10s, etc. The reproduction on FHM 229 shows a circle marked out, with Populeam virgam matre regina tenebat written in the middle. The commentary says this "problem is comparatively rare in fifteenth century texts", which doesn't seem like a fair assessment to me. Calandri. Aritmetica. c1485. Ff. 102v103v, pp. 205-207. Coloured plate opp. p.192 of the text volume. (Tropfke 654 gives this in B&W.) Franciscans and Camoldensians on a boat: 15 & 15 counted by 9s. Pacioli. De Viribus. c1500. Probs. 56-60.  x  thFf. 99r 102r. LVI. (Capitolo) de giudei Chri'ani in diversi modi et regole. a farne quanti se vole etc (Of Jews and Christians in diverse methods and rules, to make as many as one wants, etc.). = Peirani 140143. Does 2 & 30 by 9s there is a diagram for this in the margin of f. 100r, but it is not in the transcription and Peirani says another diagram is lacking. Pacioli suggests counting the passengers on shore and doing the counting out with coins or pebbles in case one will need to know the arrangement in a hurry. He also says one might count by 8s, 7s, 6s, 13s, etc., with any number of Christians and Jews. Ff. 102r 102v. [Unnumbered.] de .18. Giudei et .2. Chri'ani. = Peirani 144. 2 & 18 by 7s. F. 102v. LVII. C(apitolo). de .30. Giudei et .2. contando per .7. ch' toca va in aqua (Of 30 Jews and 2 counting by 7 with the touched going in the water). = Peirani 144. Ff. 102v 103r. LVIII. C(apitolo). de .15. Giudei et .15. Chri'ani per .9. in aqua (Of 15 Jews and 15 Christians by 9 in the water.) = Peirani 144145. 15 & 15 by 9s. In order to remember the arrangement, he says to see the next section. Ff. 103r. LIX. C(apitolo). Quater quinque. duo. unus. tres unus. et unus. bis duo. ter unus. duo duobus un' (4, 5, 2, 1, 3, 1, 1, 2, 2, 3, 1, 2, 2, 1). = Peirani 145. Pattern for 15 & 15 counted by 9s. Ff. 103r 103v. LX. C(apitolo). si da unaltro verso viz. Populea. irga. mater regina. reserra. ['unaltro' is in the margin with a mark showing where it is to go.] Vowel mnemonic for 15 & 15 counted by 9s, which he explains in detail. Agostini says this is intended to be: Populeam virgam mater regina reserrat but both Pacioli's heading and his discussion have Populea irgam mater regina reserra. The Index has LVIIILX under one heading which only refers to 'versi memorevili'.  x  tElias Levita der Deutsche. Ha-Harkabah. Rome, 1518. ??NYS. Attributes to ben Ezra, c1150??. Smith & Mikami, p. 84, say this seems to be the first printed version of the problem. Cardan. Practica Arithmetice. 1539. Chap. LXI, section 18, ff. T.iiii.r - T.v.r, but the material of interest is just a few sentences on f. T.iv.v (p. 113). Very brief description of 15 white and 15 black as 'ludus Josephus', saying one can work out any numbers with some pebbles. MUS says this is first to relate the problem to Josephus as the last man, but he doesn't give any numerical details. Hans Sachs (14941576). Meisterleid: 'Historia Die XV Christen und XV TGrcken, so auff dem meer furen'. (MUS II 132-133 gives text.) Tartaglia. General Trattato, 1556, art. 203, pp. 264v-265r. 15 whites and 15 blacks (or Turks and Christians) counted out by 3, 4, ..., 12. No reference to Josephus. Buteo. Logistica. 1559. Prob. 89, pp. 303304. 15 Christians and 15 Jews on a ship counted by 10s. [Mentioned in H&S 52.] Simon Jacob von Coburg. Ein new und WolgegrGndt Rechenbuch .... 1565 or 1612 (in quarto, not to be confused with octavo versions of 1565 and 1613 which do not contain the problem), f. 250v. ??NYS described in MUS II 133134. 12 drinkers deciding who shall pay the bill. Ahrens doesn't specify the counting number. Ahrens & Bolte (below) say this is the earliest example, after Cardan, of finding the last man. Ahrens describes numerous later examples of this type from 1693 on. Pr)vost. Clever and Pleasant Inventions. (1584), 1998. Pp. 183185. This seems like a version of the Josephus problem but isn't. Place ten counters in a circle and then ten on top of them. Start anywhere and count off five and remove the top counter. He says to count five again starting on the place where the counter was removed, so we now would say he is counting four and remove the top counter. Continue in this way, counting the places where a top counter has been removed and you manage to remove all the top counters. In fact this is impossible, but after removing five counters, you subtly start counting from the next position rather than where the top counter was removed! Hence you remove the top counters in the order 5, 9, 3, 7, 1, 6, 10, 4, 8, 2. Your audience will not observe this and hence cannot reproduce the effect. The mathematical description is simpler if one counts by fours, removing 4,8,2,6, 10, 5, 9, 3, 7, 1. The first five values are the values of 4a (mod 10) for a=1,2, 3, 4, 5. Because GCD (4, 10) = 2, this sequence repeats with period 5. Your trick shifts from the even values to the odd values and then you can count out the five odd values. Bachet. Problemes. 1612. Pr)face, 1624: A.5.v A.7.r; 1884: 89 & prob.XX,1612:103-106; prob.XXIII,1624: 174177; 1884:118-121. Turks & Christians discusses Josephus as last man. van Etten. 1624. Prob. 7 (7), pp. 7-9 (16-19). 15 Turks and 15 Christians counted by 9s. Mnemonics: Populeam virgam mater Regina ferebat; Mort tu ne failliras pas en me liurant le trespas. Discusses other cases, Roman decimation and Josephus as 40 counted by 3s. In the 1630 edition, 40 is changed to 41. Henrion's Notte, pp. 9-10, refers to Bachet's prob. 23 and mentions the correction of 40 to 41. Hunt. 1631 (1651). Pp. 266269 (258261). 15 Christians & 15 Turks counted by 9s; mnemonic: Populeam virga mater regina ferebat. Then does the same counting by 10s and gives the mnemonic: Rex Paphicum Gente Bonadat Signa Serena. Yoshida (Shichibei) Ky (= Mitsuyoshi Yoshida) (15981672). Jink-ki. Additional problems in the 2nd ed., 1634. Op. cit. in 5.D.1. ??NYS. Shimodaira (see the entry in 5.D.1) discusses the Josephus problem on pp. 1214. He gives some of the information on the Japanese names and on Michinori (11061159) and Kenk, c1331, which is presented under them. I have a transcription of (some of?) Yoshida into modern Japanese which includes this material as prob. 3 on pp. 6667. 15 children (in black) and 15 stepchildren (in white) counted by 10s. When 14 stepchildren are eliminated, the last stepchild says the arrangement was unfair and requests the counting to go the other way from him (so that he is number 1 in the counting). His stepmother agrees and thereby eliminates all her own children. This is discussed in Smith & Mikami, pp. 8084. They quote a slightly later version by Seki Kwa (16421708) where the stepmother simply reverses the order due to overconfidence. (On p. 121, they identify the source as Sandatsu Kempu, a MS of Kwa.) This is also discussed in MUS II 139-140, where it says that the change in counting was an error on the stepmother's part. Needham, p. 62, gives a picture from the 1634 ed. of Yoshida, but this is different than the picture in my modern transcription. I have a photocopy from an 1801 ed. Dean, 1997, gives the picture, discusses this and provides some additional details, citing the Heibonsha encyclopaedia for the version with the intelligent stepchild. Dean, 1997, also gives an illustration from a 1767 version called Shinpen Jinkoki, cf at 1767. Ahmed elQalyubi (d. 1659). Naouadir (or Nauadir), c1650?, published at Boulaq (a suburb of Cairo), 1892, hist. 176, p. 82. ??NYS described in Basset (18861887 below) and MUS II 136. 15 Moslems and 15 infidels on a ship counted by 9s. Muramatsu Kuday Mosei. Mantoku Jink-ri. 1665. ??NYS described in MUSII139 and Smith & Mikami, pp. 8084. Smith & Mikami, p. 81, and Dean, 1997, give Muramatsu's schematic diagrams. The top diagram is for the classic 15 and 15 counted by 10s. The second has 32 people counted by 10s to the last two, though the first 15 are coloured black and the second 15 are coloured white, with the last two drawn as squares marked by dice patterns for 5 and 6. Wingate/Kersey. 1678?. Prob. 3, pp. 531532. 15 & 15 counted by 9s or 10s or any other. Christians and Turks. From numbers, aid and art / Never will fame depart. Discusses Josephus. Thomas Hyde. Historia Nerdiludii, hoc est dicere, Trunculorum; .... (= Vol. 2 of De Ludis Orientalibus, see above for vol. 1.) From the Sheldonian Theatre (i.e. OUP), Oxford, 1694. De Ludo Char3gj seu Char3tch, pp. 225226. Says it is an Arabic game, i.e. Ludus Exeundi & Eliminadi. The description is very vague, but it seems to involve counting out in a circle. The diagram shows a circle of 21 and the text mentions counting by ten or by five. No reference to any other version of the process. ??need to read the Latin more carefully. Emil Schnippel (& Johannes Bolte). Das St. PetersSpiel (with a Nachtrag by Bolte). Zeitschrift fGr Volkskunde 39 (1929) 190192 (& 192194). Schnippel describes the appearance of solutions of the St. PetersSpiel = Sankt Pders Lek = Saint Peter's Lake on 1718C rune calendars from Sweden, which mystified academics until identified by G. Stephens in 1866. He gives the vowelmnemonic: Populeam virgam mater regina ferebat. The rune marks are X for .#%) (Xristianoi) and I for +  (Ioudaioi). He gives the German vowelmnemonic: Gott schuf den Mann in Amalek, der (or den) Israel bezwang. He cites other writers (??NYS) who describe a 1497 MS with 15 & 15 Christians and Jews counted by 10s, and versions counted by 7s and 11s, and a version with 17 & 15 Christians and Jews counted by 12s. He cites: a 1604 reference to Josephus but without specific numbers; a 1703 version with 15 & 15 French and Germans; and a 1782 version with 30 deserters, 15 to be pardoned. [Nigel Pennick; Mazes and Labyrinths; Robert Hale, London, 1990, p. 37, says that in Finland, stone labyrinths are sometimes called "Pietarinleikki (St Peter's Game). The latter name refers to a traditional numerical sequence which appears to be related to the lunar cycle. It is known from rock carvings and ancient Scandinavian calendars and as an antisemitic folktale." Can anyone provide details of a connection to the lunar cycle or its appearance in rock carvings??] Bolte's Nachtrag cites Gaidoz et al. (below at 18861887) and MUS and an article by himself in Euphorion 3 (1896) 351362, ??NYS cited MUS II 132. He sketches the history as given by Ahrens. Mentions the Japanese versions and reproduces Matuoka's picture. He adds three citations including a 1908 Indian version with 15 honest men and 15 thieves counted by 9s to the last man (??). He gives vowelmnemonics in Latin, French, German, English and Italian as follows. Non dum pena minas a te declina degeas. Populeam virgam mater regina ferebat. Mort, tu ne falliras pas. En me livrant le tr)pas. So du etwan bist gfalln hart, Stehe widr, Gnade erwart. Gott schlug den Mann in Amalek, den Israel bezwang. From member's (sic) aid and art, Never will fame depart. On tu ne dai la pace ei la rendea. Ozanam. Murphy, note 4, says the problem is not in the 1694 ed. but see below which could explain why Murphy didn't find it here. Ozanam. 1696. Preface to vol. 2 first and second of unnumbered pages, which are pp.269-270. 1708: Author's Preface second and third of unnumbered pp. Discusses Josephus, citing Bachet. Ozanam. 1725. Prob. 45, 1725: 246-250. Prob. 17, 1778: 168171; 1803: 168171; 1814:148150. Prob. 16, 1840: 7677. 15 Turks and 15 Christians counted by 9s. Gives two verse mnemonics: Mort, tu ne failliras pas, En me livrant le tr)pas; Populeam virgam mater Regina ferebat. Discusses decimation. Quotes Bachet on Josephus and asserts Hegesippus says Josephus used the method and suggests 41 counted by 3s (however, Hegesippus doesn't say this!). Kanchusen. Wakoku Chiekurabe. 1727. Pp. 8 & 35. 8 and 8 counted by 8s. This is pointing out the remarkable fact that one can count out either set first by starting at different points, in different directions. See: Dudeney, 1899 & 1905; Shaw, 1944? Minguet. 1733. Pp. 152154 (1755: 110111; 1822: 169171; 1864: 146148). 15 & 15 by 9s, whites and blacks. Populea virga pacem regina ferebat. Alberti. 1747. 'Modo di disporre 30 cose ...', pp. 132-134 (77-78). 15 Christians and 15 Turks or Jews, counted by 3, 8, 9, 10. Les Amusemens. 1749. Prob. 16, p. 138: Tir) de Josephe l'Historien. 15 and 15 counted by 9s. French and Latin mnemonics: Mort tu ne failliras pas En me livrant le tr)pas; Populeam Virgam Mater Regina ferebat. Shinpen Jinkoki (New Edition of the Jonkoki), more correctly entitled Sanpo Shinan Guruma (A Mathematical Compass). 1767. BL ORB 30/3411. ??NYS illustration reproduced in Dean, 1997. Fujita Sadasuke. Sandatsu Kaigi. 1774. ??NYS cited in a draft version of Dean, 1997, as a Japanese commentary on the problem. Hooper. Rational Recreations. Op. cit. in 4.A.1. 1774. Vol. 1, recreation XIII, pp. 4243. 30 deserters of whom 15 are to be punished, counted by 9s. Populeam virgam mater regina ferebat. L. Euler. Observationes circa novum et singulare progressionum genus. (Novi Comment. Acad. Sci. Petropol. 20 (1775 (1776)) 123-139.) = Opera Omnia (1) 7, (1923) 246-261. Gets the recurrence for the last man: L(n)  L(n1) + k (mod n). Miyake Kenry. Shojutsu Sangaku Zuye. 1795. ??NYS. (Described in MUS II 142-143.) First(?) to modify Yoshida's problem (1634?) so that the last stepchild sees his imminent fate and asks for the count to restart with him. Smith, History II 543 and Smith & Mikami, p. 82, give a poorish picture from this. Dean, 1997, is a better picture. Matuoka (= Matoka ??= Matsuoka Nichi). 1808. ??NYS translated by Le Vallois, with reproductions of the pictures, cf below. Ahrens, MUS II 140-142, discusses this, based on Le Vallois and reproduces the main picture from Le Vallois. Gives Miyake's version. Le Vallois gives the title as: Mamako tate no koto (Probl/me des beauxfils (i.e. stepsons)). There is a diagram showing the countingout processes. Ingleby. Ingleby's Whole Art of Legerdemain, containing all the Tricks and Deceptions, (Never before published) As performed by the Emperor of Conjurors, at the Minor Theatre, with copious explanations; Also, several new and astonishing Philosophical and Mathematical Experiments, with Preliminary Observations, Including directions for practicing the Slight of Hand. T. Hughes & C. Chaple, London, nd [1815]. Trick L. The Turks and Christians, pp. 104106. 15 & 15 counted by 9s. "This ingenious trick, which is scarcely known, ...." "From numbers, aid, and art, / Never will fame depart." Sanpo Chie Bukuro (A Bag of Mathematical Wisdom). 1818. BL ORB 30/3411. ??NYS illustration reproduced and discussed in Dean, 1997. Here a man and a woman are studying a set of 29 black and white go stones and the text describes the problem and how to arrange the children. Badcock. Philosophical Recreations, or, Winter Amusements. [1820]. Pp. 8385, no. 130: Thirty soldiers having deserted, so to place them in a ring, that you may save any fifteen you please, and it shall seem the effect of chance. 15 & 15 by 9s. Populeam jirgam mater Regina ferebat. (jirgam must be a misprint of virgam.) Says Josephus and 'thirty or forty of his soldiers' hid in a cave and Josephus arranged to be one of the last. Jackson. Rational Amusement. 1821. Arithmetical Puzzles.  x  thNo. 16, pp. 45 & 5456. 15 Turks and 15 Christians counted by 9s. Solution gives: From numbers, aid, and art, Never will fame depart and Populeam virgam mater regina ferebat. No. 39, pp. 910 & 6162. Decimation of a troop of 40 to be counted by 10s where to place the four ringleaders so they will be the four to be shot.  x  tRational Recreations. 1824. Feat 27, p. 103. 15 Turks and 15 Christians counted by 9s. Gives: From numbers, aid, and art, / Never will fame depart. Manuel des Sorciers. 1825. ??NX Pp. 7980, art. 40. 15 & 15 by 9s. Populeam virgam mater regina ferebat. Mort, tu ne falliras pas En me livrant au tr)pas. Says one can also do 18 & 6 by 8s, etc. Cf Gaidoz, below, col. 429. Endless Amusement II. 1826? P. 117 (misprinted 711 in 1826?): Predestination illustrated. 30 and 10 counted by 12s. Boy's Own Book.  x  thThe slighted lady. 1828: 411-412; 18282: 417418; 1829 (US): 210211; 1855:565-566; 1868: 670. 13 counted down to last person by 9s. Before 1868, it gives the survivor for 2, 3, ..., 13, counted out by 9s. The partial reprieve. 1828: 417-418; 18282: 422; 1855: 571; 1868:672673; 1881:214. 30 criminals counted by 9s to eliminate 15. Populeam virgam mater regina ferebat.  x  tNuts to Crack XIV (1845), no. 72. 21 counted by 7s to the last man. Magician's Own Book. 1857.  x  thThe fortunate ninth, pp. 221222. 15 oranges and 15 apples, counted by 9s. English mnemonics based on vowel coding. Another decimation of fruit, p. 224225. 30 apples and 10 oranges, counted by 12s in order to get the oranges first.  x  tThe Sociable. 1858.  x  thProb. 31: The puzzle of the Christians and the Turks, pp. 296 & 312314. From numbers aid and art / Never will fame depart. Mort, tu ne faillras pas / en me livrant le trepas. Populeam Virgam Mater regina ferebat. Then considers counting out 10 from 40, counting by 12s. Discusses Josephus, citing Hegesippus, and suggests counting by 3s. = Book of 500 Puzzles, 1859, prob. 31, pp. 14 & 3032. Prob. 39: The landlord tricked, pp. 298 & 316. 21 counted by 7s to the last man. =Book of 500 Puzzles, 1859, prob. 39, pp. 16 & 34. = Wehman; New Book of 200 Puzzles; 1908, p. 51.  x  tThe Secret Out. 1859. The Circle of Fourteen Cards, p. 87. This appears to be counting out all 14 cards by 6s (it says by 7s, but it takes the counted out card as one for the next stage), but it's not clear what the object is. This seems to be a corruption of an earlier version?? Indoor & Outdoor. c1859. Part II, prob. 19: The landlord tricked, p. 136. Identical to The Sociable. Boy's Own Conjuring Book. 1860.  x  thThe fortunate ninth, pp. 190-191. Identical to Magician's Own Book. Another decimation of fruit, p. 194. Identical to Magician's Own Book.  x  tVinot. 1860. Art. XXVI: De l'historien Jos/phe, pp. 5556. Gives the Josephus story and does it as counting from 41 by 3s to the last man. The Secret Out (UK). c1860. A delicate distribution, p. 12. Count 13 by 9s to the last person (different context than Leske). Mentions counting 12 by 9s. Leske. Illustriertes Spielbuch fGr Mdchen. 1864? Prob. 56430, pp. 254 & 396: Aus 12 Dreizehn machen. Count 13 by 9s to the last person. M. le Capitaine Le Vallois. Les Sciences exactes chez les Japonais. With comments by Louis de Z)linski & M. S)dillot. Congr/s International des Orientalists (= International Congress of Orientalists). CompteRendu de la premi/re session, Paris, 1873. Maisonneuve et Cie., Paris, 1874. T. 1, pp. 289-299, with comments on pp. 299303. The material of interest is on pp. 294298. Gives a translation of Matoka, 1808, and reproduces the pictures, cf above. Discusses Bachet, Ozanam, Mort tu ne failliras pas En me livrant le tr)pas, Populeam virgam mater Regina tenebat, Josephus (saying Josephus arranged to be last). Hanky Panky. 1872. The landlord tricked, pp. 129130. Identical to The Sociable, prob. 39. Kamp. Op. cit. in 5.B. 1877. No. 7, pp. 323-324. 28 counted by 9s until one is left. Footnote seems to refer to 24 counted by 9s. Mittenzwey. 1880. Prob. 282285, pp. 5052 & 100101; 1895?: 311314, pp. 5456 & 102103; 1917: 311-314, pp. 49-50 & 97-98.  x  th282 (311): 15 Negroes and 17 Europeans on a ship, counted by 12s. 283 (312): 7 students and a crafty Jew who wishes to make two of the students, A & B, pay the bill since they had been rude to him. Initially he is not included. Starting with A and counting clockwise by 3s, A & B are left. Starting with B and counting anti-clockwise by 3s, A & B are left. Then the Jew is included and starts with himself, counting anticlockwise by 3s and again A & B are left. 284 (313): 14 counted by 10s to the last man. 285 (314): 21 counted by 8s to the last man.  x  tCassell's. 1881. P. 103: To reward the favourites, and show no favouritism. = Manson, 1911, p. 256. 15 & 15 counted by 9s. Henri Gaidoz, Isra-l L)vi & Ren) Basset. Le jeu de Saint-Pierre Amusement arithm)tique. This is a series of five notes in M)lusine 3 (1886-87).  x  thGaidoz. Part I. Col. 273-274. Gives classical version with St. Peter, 15 Christians & 15Jews counted by 9s. He then gives two versions from Ceylon. One version is called Yonmaruma The massacre of the Moors and has 15 Portuguese & 15Moors with a Singhalese verse mnemonic. The second version involves the Portuguese siege of Kandy in 1821, again 15 & 15 by 9s, but different versions have the Portuguese winning or losing. These versions come from: The Orientalist 2 (1885) 177, ??NYS. The editor of The Orientalist added a version learned from an Irish soldier with the vowel-mnemonic: From number's aid and art, Never will fame depart. Gaidoz says he cannot venture a source for the puzzle. Gaidoz. Part II. Col. 307-308. Comments on correspondence generated by Part I which provided: 'Populeam virgam mater regina ferebat'; the version with the Virgin instead of St. Peter; a version with negroes and whites and a negro captain; a version with French and English; references to Josephus, Bachet and Ozanam. L)vi. Part III. Col. 332. Says ibn Ezra's "Tahboula" (Stratagem), c1150, is devoted to this game. Cites Schwenter (1623); Steinschneider's 1880 article discussed above at Ezra; Steinschneider's Catalog librorum hebr. Biblioth. Bodleianae, col. 687 all ??NYS. Previously Steinschneider opined the game derived from Jahia ibn al-Batrik's Secret of Secrets (8C), but L)vi says that that is a different amusement involving 9. Gaidoz. Part IV. Col. 429. Cites: Le Manuel des Sorciers, Paris, 2nd ed., 1802, p. 70 for a version with French and English. ??NYS, but see the 1825 ed above. Basset. Part V. Col. 528. Describes elQalyubi, c1650? cf above.  x  tRobert Harrison. UK Patent 15,105 An Improved Puzzle or Game. Applied: 25 Sep 1889; accepted: 2 Nov 1889. 2pp + 1p diagrams. 12 whites and 12 blacks on a boat with a lifeboat that will hold 12, counted by 6s, called The Captain's Dilemma. (. Ducret. R)cr)ations Math)matiques. Op. cit. in 4.A.1. 1892?  x  thPp. 105106: Une dame pas contente. 13 counted by 2s to last person. Pp. 118119: Stratag)me de Jos)phe. 41 counted by 3s to last two, claimed to be the method used by Josephus. Pp. 120121: Les marauders punis. 15 & 15 counted by 9s. Officers and soldiers to be executed. Pp. 121122: Les naufrages. Same numbers, with Turks and Christians on a boat. Mort,tu ne failliras pas, En me livrant au tr)pas. P. 122: Les (lections perfectionn)es. 36 counted by 10s want the first six chosen.  x  tHoffmann. 1893. Chap. 4, pp. 156157 & 210211 = HoffmannHordern, pp. 134135, with photo.  x  thNo. 54: Tenth man out. 15 whites and 15 blacks on a ship, counted by 10s, but first 15 get to go into the lifeboats. Photo on p. 135 shows L'Equipage Decime, with box and instructions, by Watilliaux, 18741895. No. 55: Ninth man out. Same, counted by 9s. Hoffmann cites Bachet and gives a Latin mnemonic. Photo on p. 135 shows La Question des Boches, with box having instructions on base, 19141918.  x  t(. Lucas. Problem 32. Intermed. des Math. 1 (1894) 9. n2 persons, counted by n until n-1 are left. "Probl/me dit de Caligula". E. CesarA. Solution to 32. Ibid., pp. 30-31. J. Franel. Deuxi/me r)ponse [to Problem 32]. Ibid., p. 31. Cites: Busche, CR 103, pp. 118, ??NYS. Adrien Akar. Troisi/me r)ponse [to Problem 32]. Ibid., pp. 189-190. E. Lemoine. Problem 330. Ibid, pp. 184-185. Asks for last man of n counted by p. Adrien Akar; H. Delannoy; J. Franel; C. Moreau. Independent solvers of Lemoine's problem. Ibid., 2 (1895) 120-122 & 229-230. Akar refers to Josephus, Bachet, etc. Moreau has the clearest form of the recurrence. Brandreth Puzzle Book. Brandreth's Pills (The Porous Plaster Co., NY), nd [1895]. P. 7: The tenth man out. Almost identical to Hoffmann, no. 54. No solution. Lucas. L'Arithm)tique Amusante. 1895. Pp. 1218.  x  thLe stratag/me de Jos/phe, pp. 1217. Prob. VI. 15 Christians and 15 Turks, counted by 9s. Vowel mnemonics: Mort, tu ne falliras pas, En me livrant le tr)pas!; Populeam virgam mater Regina ferebat. Discusses and quotes Bachet's 1624 Pr)face which gives the Josephus story and the idea of counting 41 by 3s. Prob. VII: Le proc)d) de Caligula, pp. 1718. 6 and 30 counted by 10s so as to count the 6 first.  x  tH. Schubert. Zw?lf Geduldspiele. 1895. P. 125. ??NYS cited by Ahrens; Mathematische Spiele; Encyklopadie article, op. cit. in 3.B; 1904. E. Busche. Ueber die Schubert'sche L?sung eines Bachet'schen Problems. Math. Annalen 47 (1896) 105-112. Clark. Mental Nuts. 1897, no. 13; 1904, no. 22. The ship's crew. 1897 has the usual 15 and 15 counted by 9s, starting with the captain, involving whites and blacks on a ship and half being thrown overboard. 1904 has 14 whites and 15 blacks and the captain must discharge 15 at a port. He joins the crew and starts counting from himself and wants to discharge the 15 blacks. P. G. Tait. On the generalization of Josephus' problem. Proc. Roy. Soc. Edin. 22 (1898) 165-168. = Collected Scientific Papers, vol. II, pp. 432-435. Says the Josephus passage is "very obscure, ..., but it obviously suggests deliberate fraud of some kind on Josephus' part." Develops a way of computing the last man. Les Bourgeois Punis. Puzzle from c1900, shown in S&B, p. 133. 8 and 2 counted by ?? to leave the 2 at the end. Dudeney. A batch of puzzles. Royal Magazine 1:3 (Jan 1899) & 1:4 (Feb 1899) 368372. The prisoners of Omdurman. 8 Europeans followed by 8 Abyssinians in a ring. Start counting with the first European. Determine the countingout number to eliminate the Abyssinians in sequence. Doing it in reverse sequence works for any multiple of 16, 15, 14, ..., 9. The LCM is 720720. But doing it in forward sequence can be done with 360361. Since the pattern is symmetric in the two types of people, a change of initial position, but keeping the same direction, will count out the others first. Dudeney. "The Captain" puzzle corner. The Captain 3:2 (May 1900) 97 & 179 & 3:4 (Jul1900) 303. The "blacks" and "whites" puzzle or The twelve schoolboys. 6consecutive "blacks" and 6 consecutive "whites" in a circle. What is the smallest number to count out by which will count out the "whites" first? You can start anywhere and in any direction. Answer is 322 and one starts counting on the fourth "white" in the direction of counting. Sreeramula Rajeswara Sarma. Mathematical literature in Telugu: An overview. Sree Venkateswara University Oriental Journal 28 (1985) 7790. Telugu is one of the Dravidian languages of south India, spoken in the area north of Madras, and is the state language of Andhra Pradesh. On pp. 8387 & 90, he reports finding examples in Telugu in the notebooks of the schoolmaster Panakalu Rayudu (18831928) who was a collector of material from many sources. Unfortunately there is no indication of where Rayudu obtained these examples and Sarma knows of no Indian versions. He has 15 thieves and 15 brahmins counted by 9s, then 30 thieves and 30 brahmins counted by 12s. Solutions are given in some literary form. The second problem is new to me. In his notes, Sarma cites the German mnemonic Gott schuf den Mann in Amalek, der (or den) Israel bezwang given by Schnippel, and that he has learned that the problem occurs in the Peddab]laiksa, a work which is unknown to me. H. D. Northrop. Popular Pastimes. 1901. No. 8: the landlord tricked, pp. 6768 & 72. = The Sociable, no. 39. Ahrens. Mathematische Spiele. Encyklopadie article, op. cit. in 3.B. 1904. Pp. 1088-1089 discusses Schubert's work and its later developments. Benson. 1904. The black and white puzzle, pp. 225-226. As in Hoffmann, no. 54, but first 15 get thrown overboard. Solution is linear rather than circular. Dudeney. Tit-Bits (14 Oct & 28 Oct 1905). ??NYS described by Ball; MRE, 5th ed., 1911, pp. 25-27. 5 and 5 arranged so that one method eliminates one group while another method eliminates the other group. Determine the two starting points and counts. Ball doesn't give these values, but seems to imply that both counts go in the same direction, and this is the case in the examples given below. Ball asks if the starting points can ever be the same for two groups of C? He gives solutions for C = 2 (counted by 4 & 3), 3 (counted by 7 & 8), 4 (counted by 9 & 5). He believes this question is new. Note on p. 27 gives the solution for C = 5, but with different starting points. See MRE, 10th ed., 1920, for a general solution with the same starting points. See: Kanchusen, 1727; Dudeney, 1899; Shaw, 1944? Pearson. 1907. Part II, no. 62, pp. 126 & 203. 15 Christians, including St. Peter, who does the counting, and 15 Jews, counted by 9s. Ball. MRE, 5th ed. 1911. See under Dudeney, 1905. Loyd. Cyclopedia. 1914. Christians and Turks, pp. 198 & 365. =MPSL2, prob. 42, pp.30-31 & 134. Like Dudeney's 1905 version with a different arrangement of 5 and 5. Williams. Home Entertainments. 1914. A decimation problem, pp. 122124. 15 whites & 15 blacks counted by 10s. Half have to go over because of shortage of provisions. Simple circular picture with man counting in middle. Ball. MRE, 10th ed., 1920, pp. 2627. See under Dudeney, 1905, for the previous version. Incorporates the solution for the case C = 5 into the text and adds a general solution due to Swinden. See: Will Blyth; Money Magic; 1926 for a related problem. Collins. Book of Puzzles. 1927. Sailors don't care puzzle, pp. 7071. 15 whites & 15 blacks counted by 10s. Captain has to throw half over because of shortage of provisions. Diagram of 15 circles in a row above a picture with 15 circles in a row below, but normally numbered it would seem natural in this problem to have the lower row numbered backward to simulate a circle. William P. Keasby. The Big Trick and Puzzle Book. Whitman Publishing, Racine, Wisconsin, 1929. Counting out problem, pp. 71 & 95. 15 & 15 shown in a circle with the starting point and direction given. Determine the counting number. Rudin. 1936. Nos. 102103, pp. 3738 & 98.  x  thNo. 102. 13 counted by 9s until last man. No. 103. 17 and 15 counted by 12s to eliminate the 15 first.  x  tErn Shaw. The Pocket Brains Trust No. 2. Op. cit. in 5.E. c1944. Prob. 50: Poser with pennies. Pattern of 5 Hs and 5 Ts given determine the count to count the Hs first, which turns out to be 11. Though not mentioned, the pattern of Hs is equivalent to that for Ts, so one can count out the Ts first by starting at a different point in the opposite direction. The pattern is the same as Dudeney (1905). Robert Gibbings. Lovely is the Lee. Dent, London, 1945. Pp. 111114. He says he was shown the puzzle by an old man on the Aran Islands. Cites Murphy, but his version is different than anything in Murphy. ??NYS information sent by Michael Behrend in an email of 12 Jun 2000. Le Rob Alasdair MacFhraing (= Robert A. Rankin, who tells me that the Gaelic particle 'Le' means 'by' and is not part of the name). ireamh muinntir Fhinn is Dhubhain, agus sgeul Iosephuis is an d! fhichead Iudhaich (The numbering of Fionn's men and Dubhan's men, and the story of Josephus and the forty Jews) (in Scots Gaelic with English summary). Proc. Roy. Irish Acad. Sect. A. 52:7 (1948) 8793. A more detailed description than in the Summary appears in Rankin's review in: Math. Reviews 10, #509b (= A995 in: Reviews in Number Theory). I had assumed that this was a development from Murphy's article, but Rankin writes that he had not heard of Murphy's article until I wrote for a reprint of Rankin's article in 1991. He gives a Scottish Gaelic version which is clearly a variant of those studied by Murphy. He then studies the problem of determining the last man, citing Tait. For counting out by 2s, the rule is simple. [There is a story that the the number of mathematicians fluent in Scots Gaelic is so small and the author's name is so obscured that the journal sent the paper to Rankin to referee, not knowing he was actually the author. The story continues that the referee made a number of suggestions for improvement which the author gratefully accepted. However, Rankin told me that he was not the referee. But he did review it for Math. Reviews!] Joseph Leeming. Games with Playing Cards Plus Tricks and Stunts. Op. cit. in 6.BE, 1949. ??NYS but two abridged versions have appeared which contain the material see 6.BE. 24 Stunts with Cards, 7th & 19th stunts. A surprising card deal & All in order. Dover: pp. 124 & 131. Gramercy: pp. 11 & 18. Both stunts involve dealing cards by putting one out face up, then the next is put at the bottom of the deck, then the next is dealt face up, .... This process is the same as counting out by 2s. The object is to produce the cards in a particular order. The first has 8 cards and wants an alternation of face cards and nonface cards. The second has the 13 cards of a suit and wants them produced in order. This is the only example that I can recall of the use of the Josephus idea as a card trick, though other forms of counting out are common, e.g. by counting 1, 2, 3, ..., or by spelling the words one, two, three, .... N. S. Mendelsohn, proposer; Roger Lessard, solver. Problem E 898 Discarding cards. AMM 57 (1950) 3435 & 488489. Basically counting out by 2s. Determine the position of the last card discarded. No mention of Josephus, though the editor asks what happens if every rth card is discarded and gets the recurrence f(N)r+f(N-1)(modN). W. J. Robinson. Note 2876: The Josephus problem. MG 44 (No. 347) (Feb 1960) 47-52. Analyses what sequences of persons can be removed by varying the count. Applies to Dudeney's problem. Barnard. 50 Observer BrainTwisters. 1962. Prob. 36: Circle of fate, pp. 41-42, 64-65 & 95. Princess counts out from 17 suitors by 3s. She sees that her favourite will be the next one out, so she reverses direction and then the favourite is the survivor. F. Jak;bczyk. On the generalized Josephus problem. Glasgow Math. J. 14 (1973) 168173. Gives a method of determining when the ith man is removed and which is the kth to be removed. Somewhat similar to Rankin's method. D. Woodhouse The extended Josephus problem. Revista Matematica HispanoAmericana 33 (1973) 207218. Gets recurrences for the last person, but unnecessarily complicates the process by considering the starting point. By combining the recurrences, he gets an n-fold iteration for the result, but this doesn't really clarify anything. Only cites Josephus. Israel N. Herstein & Irving Kaplansky. Matters Mathematical. 1974; slightly revised 2nd ed., Chelsea, NY, 1978. Chap. 3, section 5: The Josephus permutation, pp. 121128. They study the permutation where f(i) = number of ith man eliminated, but restrict to the case where one counts by 2s, which has considerable structure. Gives a substantial bibliography, mostly included here. Sandy L. Zabell. Letter [on the history of the Josephus problem]. Fibonacci Quarterly 14 (1976) 48 & 51. Sketches the history. I. M. Richards. The Josephus problem. MS 24 (1991/92) 97104. Studies the case of counting out by 3s. Shows the 'Tait numbers', i.e. n such that L(n) = 1 or 2, are given by [(3/2)i + 1/3], where = 1.216703..., and obtains a formula for L(n). Presumably this could be extended to the general case?? Michael Dean. Josephus and the Mamakodate san (Scheme to benefit the stepchildren). International Netsuke Society Journal 17:2 (Summer 1997) 4153. There are inro boxes from late 17C Japan which have pictures of the 15 children and 15 stepchildren problem. These initially mystified the art historians, but eventually they discovered the Josephus problem and its Japanese forms, but only as far back as Bachet. Dean gives a brief history for the benefit of art collectors, with references to a number of Japanese sources (some of which I have not seen) see above at 1767, 1795, 1818 and some photos of the inro boxes (including a fine late 17C example from the collection of Michael and Hiroko Dean) and other material. Ian M. Richards. The Josephus Problem and Ahrens arrays. MS 31:2 (1998/9) 3033. He has finally obtained a copy of Ahrens' work, but from the first edition, and states the result clearly. For n persons, labelled 1, 2, ..., n, counted out by k, if we want to locate the eth man counted out, form a sequence starting with 1 + k(ne) and then form each next term by multiplying a term by k/(k1) and rounding the result up to an integer. (I.e. xn+1 = Ixn * k/(k1)J.) Then the position number of the eth person eliminated is the difference between kn + 1 and the largest term in the sequence less than kn + 1. The sequence is giving the points where L(n, k) is zero in some sense. Note that when e=n, so we are looking for the last person, then the sequence starts at 1, which is because we start counting with the first person as one. kn + 1 is the total amount counted in counting n people by k, For other values of e, the change of the starting point of the sequence compensates for the fact that one only counts k(ne) + 1 to eliminate the eth person. Ahrens then examined the sequences obtained, with rational multipliers, and found some nice properties which Richards states. Richards generalises to arbitrary multipliers and finds connections with Beatty sequences, KanL. Ian M. Richards. Towards an analytic solution of the Josephus problem. Unpublished preprint sent to me on 21 Mar 1999, 12pp. (Available from the author, 3 Empress Avenue, Penzance, Cornwall, TR18 2UQ.) Gets formulae for the case k = 4 which give the result with a maximum error of 1. David Singmaster. Adjacent survivors in the Josephus Problem. Nov 2003, 5pp, but may be extended. This was inspired by the first example in Pacioli's De Viribus, which has 2 'good guys' and 30 'bad guys' arranged in a circle and every 9th person is thrown overboard. I was struck by the fact that the two survivors were adjacent in the original circle as clearly marked in the marginal diagram. Offhand it seems an unlikely result, but one soon observes that this remains true as the counting out takes place. That is, if the two survivors in counting off N by Ks are adjacent, then this is also true for counting off n by Ks for 3  n < N. This paper investigates the maximal N for which counting out by Ks leaves two last survivors who were originally adjacent. 7.C.EGYPTIAN FRACTIONS  tThe basic problem is to represent a given fraction as a sum of fractions with unit numerators and distinct denominators, as done by the Egyptians. NOTE: Dating of early Egyptian documents is rather uncertain and sources can vary by several hundred years. I will tend to use dates of Neugebauer and Parker, as given in Gillings. This dates the composition of the Rhind Papyrus and the Moscow Papyrus as 13th Dynasty, c-1785, but other sources say the Moscow Papyrus is several hundred years older and other sources date the composition of the Rhind Papyrus to the 12th Dynasty, c1825.  t Papyrus Rhind, composed c1785 (or c1825), copied c-1650 (or c1700). A. B. Chace, ed. (1927-29); c=NCTM, 1978. Pp. 21-22, 50-51. Moscow Mathematical Papyrus. c1785. W. W. Struve, ed; Mathematischer Papyrus des Staatlichen Museums der Sch?nen KGnste in Moskau; Quellen und Studien zur Geschichte der Mathematik, Abt. A: Quellen, Band 1; Springer, 1930. Fibonacci. 1202. Pp. 77-83 (S: 119126): ... de disgregatione partium in singulis partibus [... on the separation of fractions into unit fractions]. He clearly has the idea of taking the smallest n such that 1/n  a/b, but he doesn't prove that this gives a finite sequence. J. J. Sylvester. On a point in the theory of vulgar fractions. Amer. J. Math. 3 (1880) 332-335 & 388-389. M. N. Bleicher. A new algorithm for the expansion of Egyptian fractions. J. Number Theory 4 (1972) 342-382. The Introduction, pp. 342-344, outlines the history. Pp. 381-382 give 41 references. E. J. Barbeau. Expressing one as a sum of distinct reciprocals. CM 3:7 (1977) 178-181. Bibliography of 20 items. Paul J. Campbell. A "practical" approach to Egyptian fractions. JRM 10 (197778) 8186. Discusses Fibonacci & Sylvester's methods, etc. 22 references. Charles S. Rees. Egyptian fractions. Math. Chronicle 10 (1981) 13-30. Survey with 47 references. R. J. Gillings. Mathematics in the Time of the Pharaohs. Dover, 1982. He has a long discussion on the Egyptian approach to this topic, discussing and comparing the work in the various sources: Reisner Papyri (c2134); Rhind Papyrus (c1785); Moscow Papyrus (c1785); Kahun Papyri (c1785, but later than the previous two items); Egyptian Mathematical Leather Roll (c1647), but he certainly devotes most space to the Rhind Papyrus and the Leather Roll. 7.D.THE FIRST DIGIT PROBLEM S. Newcomb. Note on the frequency of use of the different digits in natural numbers. Amer. J. Math. 4 (1881) 39-40. Obtains the law by simply considering logarithms. F. Benford. The law of anomalous numbers. Proc. Amer. Phil Soc. 78 (1938) 551-572. E. H. Neville. Note 2540: On even distribution of numbers. MG 39 (No.329) (Sep 1955) 224-225. Says the problem is not precisely defined. (Not cited in Raimi.) R. A. Fairthorne. Note 2541: On digital distribution. Ibid., p. 225. Cites earlier results (see Raimi) and says the law is "a consequence of the way we talk about [numbers]." (Not cited in Raimi.) R. A. Raimi. The first digit problem. AMM 83 (1976) 521-538. Extensive survey and references. G. T. Q. Hoare & E. E. Wright. Note 70.5: The distribution of first significant digits. MG 70 (No. 451) (Mar 1986) 34-37. Generates numbers as ratios of reals uniformly distributed on (0, 1). Finds explicit and surprisingly simple probabilities for initial digits of these numbers, which are reasonably close to Benford's probabilities. Peter R. Turner. The distribution of l.s.d. and its implications for computer design. MG 71 (No. 455) (Mar 1987) 26-31. l.s.d. = leading significant digit. Cites some recent articles. 7.E. MONKEY AND COCONUTS PROBLEMS  tMost of these problems are determinate. Mahavira gives two indeterminate problems, but the next are in Ozanam, with the classic version of the problem first reappearing in Carroll,1888; Ball, 1890; Clark, 1904; and Pearson, 1907, qv. NOTATION. The classic coconuts problem has the following recurrence for the number of coconuts remaining: Ai+1 = (n1)/n [Ai 1], i.e. each sailor removes 1 (given to the monkey) and 1/n of the rest. There are two common endings of the problem.  x  thEnding 0 the n-th man leaves a multiple of n, so the monkey doesn't get a final coconut. See: Mahavira: 131, 132; Williams; Moritz; Meynell; Leeming. Ending 1 the n-th man leaves one more than a multiple of n, so the monkey gets another coconut. See: CarrollWakeling; Ball; Clark; Pearson; Roray; Collins; Kraitchik; Phillips; Home Book; Leeming; Devi; Allen. One can extend this to Ending E the nth man leaves a number  E (mod n).  x  tOther indeterminate versions: Ozanam; Dudeney; Weber (Dirac); Rudin. For the solution with (n1) coconuts, see: Roray; Weber (Dirac); Birkhoff & Mac Lane; Anonymous in Eureka; Gardner; Pedoe, Shima & Salvatore; Singmaster. See Morris (1988); Singmaster (1993) for the alternative division form where the pile is divided equally and the monkey takes one from the remainder, i.e. each sailor takes 1/n of the pile and then the monkey then takes 1 from the remainder, so the recurrence is Ai+1Ġ=(n-1)Ai/n 1. This is similar to the form of recurrence occurring in the determinate versions of the problem, where division takes place first and then some more is included. Comparing this with the standard form, we see that the forms can be described by the number of coconuts (mod n) at each stage. In the classic form, each Ai  1 (mod n), and in Morris's form, each Ai  0 (mod n), so we can conveniently name these Form 1 and Form 0. Unless specified, all examples have Form 1. It is easy to generalize to giving c coconuts to the monkey at each stage, in either Form, which I call Forms 1c and 0c, but only Anonymous in Eureka; Kircher; Pedoe, Shima & Salvatore; Singmaster consider this. Only Kircher considers giving variable amounts to the monkey and he even permits negative values, e.g. if the monkey is adding coconuts to the pile! Birkhoff & Mac Lane; Herwitz; Pedoe, Shima & Salvatore consider a variation where no ending is specified except that there is an integral number left after the nth division. A discussion of this version has now been added to Singmaster. Jackson gives a simple form with no monkey. Edwards gives a form where the monkey only gets a coconut at the end. See Tropfke 582. See also 7.S.1. Hermelink, op. cit. in 3.A, says there are Egyptian versions, presumably meaning some of the simpler determinate types of heap or 'aha' problems in the Rhind Papyrus. Old Babylonian tablet YBC 4652. c1700?. Transcribed, translated and commented on in: O.Neugebauer & A. Sachs; Mathematical Cuneiform Texts; American Oriental Society and American Schools of Oriental Research, New Haven, 1945, pp. 100103, plate 13 & photo on plate 39. This has fragments of 22 simple problems, of which six can be restored. The authors say the dating of the tablets discussed in the book is quite uncertain, only stating "they are to be dated to the centuries around 1700 B.C."  x  thNo. 7 is reconstructed as: I found a stone, but did not weigh it; after I added oneseventh and added oneeleventh, I weighed it: 1 mana. What was the original weight of the stone? In modern notation, this is: x + x/7 + (x + x/7) / 11 = 1, or simply: x (8/7) (12/11) = 1 which is a simple 'aha' problem. No. 8 leads to x x/7 + (x x/7) / 11 = 1. No. 9 leads to x x/7 + (x x/7) / 11 [x x/7 + (x x/7)/11] / 13 = 1. No. 19 leads to 6x + 2 + (6x + 2)24/21 = 1. No. 20 leads to 8x + 3 + (8x + 3)21/39 = 1. No. 21 leads to x x/6 + (x x/6) / 24 = 1.  x  tOld Babylonian tablet YBC 4669. c1700?. Neugebauer and Sachs continue on p. 103 with a new analysis of this table which Neugebauer had previously treated in Mathematische Keilschrifttexte III, op. cit. in 6.BF.2, p. 27. It leads to (2/3) (2/3) x + 10 = x/2. Chiu Chang Suan Ching (Jiu Zhang Suan Shu). c-150.  x  thChap. VI, prob. 27, p. 69. Man carrying rice through customs pays 1/3, then 1/5, then 1/7 and has 5 left. Chap. VI, prob. 28, pp. 69-70. Man pays 1/2, 1/3, 1/4, 1/5, 1/6, making 1 paid out. Chap. VII, prob. 20, pp. 79-80. Man gains 30% and sends home 14000; then gains 30% and sends 13000; then 30% and 12000; then 30% and 11000; then 30% and 10000; leaving 0. Capital was 30468 84876/371293. (English in Lam & Shen, HM 16 (1989) 113.)  x  tZhang Qiujian (= Chang Chhiu-Chien = Chang Ch'iu Chien = Zhang Yo Chien). Zhang Qiujian Suan Jing (= Chang Chhiu-Chien Suan Ching) (Zhang Qiujian's Mathematical Manual). 468. ??NYS. Chap. II, no. 17. Man gains 40% and withdraws 16000; then gains 40% and withdraws 17000; then gains 40% and withdraws 18000; then gains 40% and withdraws 19000; then gains 40% and withdraws 2000; leaving 0. Capital was 35326 5918/16807. (English in Lam & Shen, HM 16 (1989) 117.) Anania Schirakatzi (= Ananias of Shirak). Arithmetical problems. c640. Translated by: P. Sahak Kokian as: Des Anania von Schirak arithmetische Aufgaben; Zeitschrift fGr d. deutsch?sterr. Gymnasien 69 (1919) 112117. Kokian cites several versions and editions of this Armenian MS as well as some studies on Ananias, but I haven't been able to determine just where Shirak was. The title varies on the different MSS and Kokian heads the text with one version translated into German: Des Anania Vardapet Schirakuni Frage und Aufl?sung [Questions and Solutions of the Priest Ananias of Shirak.] There are 24 problems, mostly of the 'aha' or 'heap' type. Only the numerical solutions are given no methods are given. There are several confusing errors which may be misprints or may be errors in the MS, but Kokian says nothing about them. One problem seems to have omitted an essential datum of the number of grains of barley in a 'kaith'. I cannot reconcile one solution with its problem (see 7.H).  x  thProb. 11. Spend 5/6 thrice, leaving 11. Answer: 2376. Prob. 13. Spend 3/4 thrice, leaving 5. Answer: 320. Prob. 19. (Double and give away 25) thrice to leave zero. Answer: 21E. Kokian notes that this and prob. 22 are the earliest occurrences of fraction signs in Armenian. Hermelink, op. cit. in 3.A, points out that here the doubling is done by God in response to prayer in churches then the Arabic world converts the churches to mosques, and then the West reverts to churches, while in the Renaissance, the doubling is by winning at gambling. In fact, during the Renaissance, it often was by profit from trade. Prob. 21. Give away 1/2, then 1/7, then 1/8, then 1/14, then 1/13, then 1/9, then 1/16, then 1/20, leaving 570. Answer: 2240.  x  tPapyrus of Akhmim. c7C. Jules Baillet, ed. Le Papyrus Math)matique d'Akhm3m. M)moires publi)s par les membres de la Mission Arch)ologique Fran'ais au Caire, vol. IX, part 1, (1892) 1-89. Brief discussion of the following problems on pp. 5859.  x  thProb. 13, p. 70. Take 1/13th, then 1/17th of the rest, leaving 150. Answer:172+1/2+ 1/8 + 1/48 + 1/96. (Also given in HGM II 544. Kaye I 48, op. cit. under Bakhshali MS, discusses the Akhmim problem and says both Heath and Cantor give misleading references, but I don't see what he means.) Prob. 17, p. 72. Take 1/17th, then 1/19th of the rest, leaving 200. Answer: 224+1/4 + 1/18.  x  tBakhshali MS. c7C. In: G. R. Kaye, The Bakhsh]li manuscript. J. Asiatic Soc. Bengal (2) 8:9 (Sep 1912) 349-361. P. 358: Sutra 25: example merchant pays customs of 1/3,1/4, 1/5 and finds he has paid 24. = Kaye III 205, f. 14r. Hoernle, 1888, op. cit. under Bakhshali MS, p. 277 gives the above and the following. Merchant gains 1/3, 1/4, 1/5, 1/6 and finds he has gained 40. (Kaye III 205, f. 14r gives this in less detail and it is not clear if Hoernle's statement is what is intended.) Merchant loses 1/3, 1/4, 1/5 for a total loss of 27. (Kaye III 205, f. 14v says the remainder is 27 but gives the original amount as 45, so he seems to have loss and remainder interchanged.). Merchant loses 1/3, 1/4, 1/5 leaving 20 (can't find in Kaye III ??). Kaye I 48, section 89, says there are 17 examples of this general form, some with the initial value given and the final result wanted, others with the final result given and the initial value wanted. Gives the first example above and two others with the same rates and a payment of 280 (Kaye III 165, ff. 52r52v) or a result of 2x 32, where x is the initial value (Kaye III 207, f. 15r). Kaye III 204, f. 13v: start with 60, lose 1/2, gain 1/3, lose 1/4, gain 1/5. Kaye III 208, f. 16r: give 2/3, then 2/5, then 2/7, then 2/9, leaving 3. How much was given? See also Datta, op. cit. under Bakhshali MS, pp. 44 & 52-53. He says the Akhmim problems give the remainder, while the Bakhshali MS and Mahavira problems give the amount paid but above we have seen both kinds. Datta, p. 46, says (Kaye III 184,) f.70v has a badly damaged problem about a king who gives away 1/2, 1/3 and 1/4 of his money, making 65 given away. Datta says that the king had only 60 to start!! But if this is a problem of the type being treated here, then the fractions are applied to the amount left after the previous stage and the king would have 1/4 of his original amount left and he must have had 861/3 to start. Ripley's Puzzles and Games. 1966. P. 78 asserts that Premysl of Staditze won the kingdom of Bohemia by solving the following. Give (half and one more) twice, then half and three more to leave zero. Typically Ripley's gives no details. The Encyclop%dia Britannica says the origin of the Premysl dynasty is obscure, deriving from a plowman who married the Princess Libuse, but giving no date, though apparently by the 9C. [Rob Humphreys; Prague The Rough Guide; The Rough Guides, London, (1992), 3rd ed, 1998, p. 249] gives the legends of the founding of Prague. The maiden queen Libue, in the 7C or 8C, fell into a trance and told her followers to seek a ploughman with two oxen. Such a man, named Pemysl (meaning ploughman) was found and produced the dynasty. He makes no mention of the problem, nor does the Blue Guide for Prague. Mahavira. 850. Chap. III, v. 129140, pp. 6769 are simple problems of this general type, involving sums of numbers diminished by fractions I give just some examples. Chap.VI, v. 112, 114, 130, 131, 132, pp. 116 & 123-125. Chap. III.  x  th133. x(3/4)(4/5)(5/6) + y(1/2)(5/6)(4/5) + z(3/5)(3/4)(5/6) = 1/2. This reduces to: x/2+ y/3 + 3z/8 = 1/2 and he arbitrarily picks two of the values, getting: 1/3,1/4, 2/3. 134. x(1/2)(5/6)(4/5)(7/8)(6/7) = 1/6.  x  t Chap. VI.  x  th112. Double and subtract 5, triple and subtract 5, ..., quintuple and subtract 5, leaving 0 (Datta & Singh I 234, note this gives 43/12 flowers and hence replace 5 by 60 to give 43). 114. Less regular problem, leaving 0. 130. Gives a general technique. 131. Two sons and mangoes (subtract 1 and halve) twice, leaving some even number i.e. Ending 0 with 2 men. Cf Pearson, 1907. 132. Man placing flowers in a temple (subtract 1 and delete @) thrice, leaving some multiple of 3 i.e. Ending 0 with 3 men. Cf Pearson, 1907. There are some simpler problems in Chap. IV, v. 29-32, pp. 74-75.  x  tChaturveda. 860. There are some simple examples on pp. 282-283 of Colebrooke. Sridhara. c900. V. 74(i), ex. 97, pp. 59-60 & 96. Give away 1/2, then 2/3, then 3/4, then 4/5, leaving 3. Tabari. Mift]h almu]mal]t. c1075. ??NYS.  x  thPp. 177f. & 128, no. 28 & 45. Tropfke 585 says these are business trips. P. 127, no. 44. Hermelink, op. cit. in 3.A, says this is Fibonacci's seven gate problem of p. 278, with oranges instead of apples. Tropfke 585 says it is a problem with porters at an orange orchard.  x  tAbraham. Liber augmenti et diminutionis. Translated from Arabic in 12C (Tropfke 662 says early 14C). Given in: G. Libri; Histoire des Sciences Math)matiques en Italie; vol. 1, pp. 304376, Paris, 1838. ??NYR cited by Hermelink, op. cit. in 3.A. Bhaskara II. Bijaganita. 1150. Chap. 4, v. 114. In Colebrooke, pp. 196-197. (Lose 10, double, lose 20) thrice to triple original. Fibonacci. 1202. Pp. 258-267, 278, 313-318 & 329 (S: 372383, 397398, 439445, 460461) gives many versions! Chap. 12, part 6, pp. 258-267 (S: 372383): De viagiorum propositionibus, atque eorum similium [On problems of travellers and also similar problems] is devoted to such problems.  x  thP. 258 (S: 372373). (Double & spend 12) thrice to leave 0 or 9. Answers:10,11D. H&S 60 gives English. P. 259 (S: 374). Start with 10, (double and spend x) thrice to leave 0. H&S 60 gives English. P. 259 (S: 374). Same, starting with 11D and leaving 9. P. 259 (S: 374375). (Triple and spend 18) four times to leave 0. Answer: 88/9. P. 260 (S: 375). Start with 8 8/9, (triple and spend x) four times to leave 0. Pp. 260-261 (S: 375376). (Triple and spend 18) four times to leave 12, or the original amount, or original amount + 20. P. 261 (S: 376377). Three voyages, making profits of 1/2, 1/4, 1/6 and spending 15 each time to leave final profit of 1/2. Answer: 24 6/7. Same, with initial amount 24 6/7, find the common expenditure. Same, with 'leave final profit 1/2' replaced by 'leave 21'. Pp. 261-266 (S: 377383). Many variations. Pp. 266-267 (S: 383). Start with 13, (double and spend 14) X times to leave 0. H&S 60 gives English. He gets X = 3 voyages, by linear interpolation between 3 and 4. Exact answer is log2Ġ14=3.80735. P. 278 (S: 397398). De illo qui intravit in viridario pro pomis collegendis [On him who went into the pleasure garden to collect apples]. Man collects apples in a garden with 7 gates. (Subtract half and one more) seven times to leave 1. H&S 60 and Sanford 221 give English. Answer: 382. Pp. 313-316 (S: 439443). Man starts with 100 and spends 1/10 twelve times. This is not strictly of the type we are looking at, but it is notable that he computes 100(.9)12 using a form of decimal fraction, getting 28.2429536481. See 7.L for related problems. Pp. 316-318 (S: 443445). Exit from a city with 10 gates. He pays 2/3 of his money and 2/3 more, then 1/i of his money and 1/i more for i=2, ..., 10, leaving 1. P. 329 (S: 460). Same as on p. 258, done by false position. P. 329 (S: 460461). Start with 12, (double and spend x) thrice to leave 0.  x  tAbbot Albert. c1240. Prob. 8, p. 334. (Double and subtract 1) thrice, leaving 0. Chu Shih-Chieh (= Zhu Shijie). Ssu Yuan YG Chien (= Siyuan Yujian) (Precious Mirror of the four Elements = Jade Mirror of the Four Unknowns). 1303. Questions in Verse, prob. 4. ??NYS. English in Li & Du, p. 179. (Double and drink 19) four times to leave 0. BR. c1305.  x  thNo. 89, pp. 108-109. (Double and spend 35) thrice leaving 0. No. 119, pp. 134-135. (Double and spend 40) thrice leaving 0.  x  tFolkerts. Aufgabensammlungen. 1315C.  x  th(Double and give a) n times to leave nothing. 17 sources. Folkerts notes the solution is a a/2n and the MSS give a general rule. (Give half and one more ) n times to leave c. 12 sources, with the MSS giving a general rule. Two sources where half is replaced by a quarter. One irregular example: Lose half, gain 2; lose half, gain 4; lose half, gain 6; to leave 4. Munich 14684, XXXIV. 6 sources. Cites a number of other sources, almost all cited in this section (two items are NYR).  x  tGherardi?. Liber habaci. c1310. Pp. 144-145. Three porters i-th takes half plus i, leaving none. Gherardi. Libro di Ragioni. 1328.  x  thPp. 47-48. Man gathering apples. Four porters i-th takes half plus 5 - i, leaving 1. P. 100. Man makes 12d on his first trip. He earns at the same rate on his second trip and then has 100d. This leads to a quadratic and he finds the positive solution. See Van Egmond, op. cit. in Common References, pp. 168, 177 & 185 for Italian, English and algebraic versions and some corrections.  x  tLucca 1754. c1330.  x  thFf. 26r-27r, pp. 64-65. Multiply by 6/5 and spend 12, multiply by 5/3 and spend 17, double and spend 20, leaving 0. F. 59v, p. 135. (Double and spend 12) thrice to leave 3.  x  tPaolo dell'Abbaco. Trattato di Tutta l'Arte dell'Abacho. 1339. The first work in the codex Plimpton 167 in the Plimpton collection, Columbia University, New York, is a c1445 copy. ??NYS described in Rara, 435-440 and Van Egmond's Catalog 254255. Van Egmond 365 lists 9 MSS of this work. MSB2433, Biblioteca Universitaria, Bologna, is a c1513 copy of just the problems of this work Dario Uri has kindly sent a copy of this, but it is somewhat blurry and often illegible; he has now sent a version on a CD which is clearer. It is dated 1339. See: Van Egmond's Catalog 6768. Rara 438 calls it the Dagomari Manuscript and reproduces a figure of a garden with three gates and guards. Only the first line of the text of the problem is included, but the text is on ff. 25r25v of B 2433. (Halve and subtract 1) thrice to leave 3. Munich 14684. 14C.  x  thProb. V, p. 78. (Double and subtract 2) some times to leave 0 determines initial values for various numbers of times as 2(1 1/2n). The text seems to also consider (Double and subtract 5). Prob. VI, p. 78. (Halve and subtract 1) thrice to leave 3. Prob. XXXIV, p. 84. (Double and subtract 100) thrice, then (double and subtract 50) thrice, leaving 0. Answer: 92 31/32,  x  tBartoli. Memoriale. c1420. Prob. 9, f. 75v (= Sesiano pp. 138 & 148). Man going into a garden to get apples. Gives 3/4 plus 3 more; 2/3 plus 2 more; 1/2 plus 1 more; to leave 1. Proven'ale Arithm)tique. Written (or more likely copied) at Pamiers, c1430. MS in Biblioth/que Nationale, Paris, fonds fran'ais, nouvelle acquisition 4140. Previously in the collections of Colbert (no. 5194) and the King (no. 7937). Partially transcribed/translated and annotated by Jacques Sesiano; Une Arithm)tique m)di)vale en langue proven'ale; Centaurus 27 (1984) 2675. The problems are not numbered, so I will give the folios and the pages in Sesiano. However the indications of the original folios have not come through on a few pages of my copy and I then only give Sesiano's page.  x  thP. 58. Man doubles his money and spends 1, triples and spends 2, quadruples and spends 2, leaving him with 3. F. 113v114r, p. 60. (Sell 1/2 and one (or 1/2 ??) more) three times to leave 3. The author gives a general solution as starting with the final result, (adding the extra number and double) three times to get the original number.  x  tPseudodell'Abbaco. c1440.  x  thProb. 47, p. 44 with plate on p.45. (Halve and subtract 2) thrice to leave 7. Prob. 71, pp. 65-67 with plate on p. 66. (Lose @ and 6 more) thrice to leave 24. (The illustrations are very different from that in Rara (see previous entry). Rara does not show enough text to see if the numbers used are the same as here, though the wording is clearly different.) I have a colour slide of this.  x  tAR. c1450. No. 185 & 187, pp. 87-88, 173-174 & 220.  x  th185 = Fibonacci, p. 258. 187. Double and spend 6, double and spend 12, double and spend 15, leaving the initial amount.  x  tMuscarello. 1478.  x  thFf. 78v79r, pp. 194196. Lose 1/2 and 6 i more, for i = 1, 2, 3, 4, 5, leaving 1. Ff. 84r85r, pp. 201204. Merchant starts with 79 and makes profits of 17%, 19%, 21%, 23% at four fairs.  x  tdella Francesca. Trattato. c1480.  x  thF. 23r (7374). Gain 1/3 + 1/4 and 20 more. Then spend 1/4 + 1/5 and 20 more to leave 24. F. 37v (9798). Double and spend 11, triple and spend 47, double and spend 34, double and spend 16 to leave 0. English in Jayawardene. F. 41v (104). Identical to f. 23r.  x  tChuquet. 1484.  x  thProb. 30. (Double and subtract 12) thrice, leaving 0. English in FHM 206. Prob. 3133 are generalized versions. E.g. Prob. 31 is double and spend 5, triple and spend 9, quadruple and spend 12 to leave 8. Prob. 95, English in FHM 219. Merchant makes a profit of 1/3 and i more on his ith journey. He makes as many journeys as he has money to start with. When does he have 15? This gives a messy equation: (4/3)x = 3 9/(x+12). Chuquet uses some interpolation to estimate X = (50 16297/16384) 4 13/128 [FHM misprints this] = 3.03949414, but I get 3.045827298. Chuquet says ordinary interpolation is not valid.  x  tCalandri. Arimethrica. 1491.  x  thF. 66v. (Double and spend 2) thrice leaving 0. F. 74r. Double and then gain 50% giving 1000. Woodcut of merchant on horse.  x  tPacioli. Summa. 1494.  x  thF. 105v, prob. 20. (Give half and one more) thrice leaving 1. (See also H&S 58.) F. 105v, prob. 22. (Double and spend 12) thrice leaving 0. F. 187r, prob. 8. Start with x and double x times to get 30. This gives us x 2x = 30, whose answer is 3.21988.... He interpolates both factors linearly on the third day, getting (3+y)(8+8y) = 30, so 3+y = 1 + (19/4) = 3.17945.... He approaches the following problems similarly. Ff. 187r187v, prob. 9. Start with x and make 25% on each of x trips to make 40% overall. This gives x (1.25)x = 1.4 x. F. 187v, prob. 10. Leads to x (1.2)x = x2. F. 187v, prob. 11. Leads to x (1.4)x = 6x. F. 188r, prob. 13. Start with 13, (double and spend 14) x times to leave 0. He observes that each iteration doubles the distance from 14, so the problem leads to 2x = 14, but again he has to interpolate on the third day.  x  tCalandri, Raccolta. c1495. Prob. 16, pp. 17-18. Merchant gains @ of his money plus i on the ith trip. After three trips he has 15. Pacioli. De Viribus. c1500. Ff. 120r 120v, 111r 111v (some pages are misbound here). C(apitolo). LXVII. un signore ch' manda un servo a coglier pome o ver rose in un giardino (A master who sends a servant to gather apples or roses in a garden). = Peirani 156158. (Lose half and one more) three times to leave 1. Discusses the problem in general and also does (Lose half and one more) five times to leave 1; (Lose half and one more) three times to leave 3. Blasius. 1513. Ff. F.iii.r F.iii.v: Decimaquinta regula. Sack of money. First man takes half and returns 100; second takes half and returns 50; third takes half and returns 25; leaving 100 in the sack. Johannes K?bel. Rechenbiechlein auf den linien mit Rechenpfeningen. Augsburg, 1514. With several variant titles, Oppenheim, 1518; Frankfort, 1531, 1537, 1564. 1564 ed., f.89r, ??NYS. Lose half, gain 100, lose half, gain 50, lose half, gain 25 to yield 100. (H&S 58-59 gives German and English.) Ghaligai. Practica D'Arithmetica. 1521. Prob. 29, f. 66v. Start with 100 and have to bribe ten guards with 1/10 each time. Computes the exact residue, i.e. 100 x .910. (H&S 5960). Tonstall. De Arte Supputandi. 1522.  x  thQuest. 43, p. 173. Give half plus i+1, for i = 1, 2, 3, leaving 1. (H&S 61 cites this to the 1529 ed., f. 103) Quest. 44, pp. 173174. Give half and get back 2i for i=1,2,3, leaving 12. P. 246. (Double and spend 12) thrice to leave 0.  x  tRiese. Die Coss. 1524. Several examples no. 35, 53, 55, 56, 58, 59, 60, 61, 62. I describe a few.  x  thNo. 35, p. 45. Man stealing apples: (give half and one more) four times, leaving 1. No. 53, pp. 47-48. (Double and spend 12) thrice, leaving 0. No. 55, p. 48. (Double and spend i) for i = 1, 2, 3, leaving 10. No. 58, p. 48. x + (4x+1) + (3(4x+1)+3) = 56. No. 61, p. 48. (Give half plus 2+2i more) for i = 1, 2, 3, leaving 0. No. 62, p. 49. (Give half less 2i) for i = 1, 2, 3, leaving 12.  x  tTartaglia. General Trattato. 1556. Book 12, art. 34, p. 199v. Book 16, art. 47 & 113116, pp.246r & 253v-254r. Book 17, art. 9 & 20, p. 268v & 271r. Final remainder specified in each case.  x  th1234. (Take half plus i more) for i = 1, 2, 3, 4, leaving 1. Cf 16115. 1647. Take 1/2 and 1 more, 1/3 and 2 more, 1/4 and 4 more, leaving 26. 16113. (Double and subtract 20) thrice, leaving 0 (H&S 61 gives Latin and English of this one and says it appears in van der Hoecke (1537), Stifel (quoting Cardan) (1544), Trenchant (1566) and Baker (1568) (but see below). 16114. (Halve and subtract 1) thrice, leaving 1, 2, .... 16115. Halve and subtract 1, then 2, 3, 4, leaving 1. Cf 1234. 16116. Lose 1/2 and 3 more, lose 2/3 and get back 10, lose 3/4 and 6 more, lose 4/5 and get back 16, leaving 24. 179. Double & spend 4, double & spend 8, leaving 24. 1720. Double and spend 18, double and spend 24, double and spend 36, leaving 280.  x  tButeo. Logistica. 1559.  x  thProb. 6, pp. 334335. Lose 1/2 and 3 more, lose 1/3 and 4 more, lose 1/4 and gain 1, to leave 100. Prob. 13, pp. 342343. Start with X, gain 40. Make the same rate of profit twice again and then the second of these gains is 90. Prob. 19, p. 347. Double and spend 12, triple and spend 15, quadruple and spend 14, leaving 12. Prob. 20, pp. 347348. Gain 1/4 and spend 7, gain 1/3 and spend 10, lose 3/7 and spend 8, leaving 0. Prob. 21, pp. 348350. (Double and spend 10) X times to leave 0. He makes an error at X = 8 and deduces X = 7.  x  tBaker. Well Spring of Sciences. 1562?  x  thProb. 7, 1580?: ff. 192r193r; 1646: pp. 302-304; 1670: pp. 344345. Lose half and gain 12, lose half and gain 7, lose half and gain 4, leaving 20. Prob. 8, 1580?: ff. 193r193v; 1646: p. 304; 1670: p. 345. (Double and spend 10) thrice leaving 12.  x  tGori. Libro di arimetricha. 1571. F. 72r (pp. 77-78). (Lose half and one more) four times to leave 3. Book of Merry Riddles. 1629? (Take half and half more) thrice, leaving one. Wells. 1698. No. 118, p. 209. Soldiers take half of a flock of sheep and half a sheep more, thrice, leaving 20. Ozanam. 1725. Prob. 28, question 1, 1725: 211-212. (Give half the eggs and half an egg) thrice. He doesn't specify the remainder and says that 8n-1 eggs will leave n-1 and that one can replace 8 by 2k if one does the process k times. Montucla replaces this by some determinate problems see below. Simpson. Algebra. 1745. Section XI (misprinted IX in 1790), prob. XV, pp. 8687 (1790: prob. XXVII, pp. 8889. Shepherd loses (half and 1/2 more) thrice to leave 5 (or 1) sheep. Les Amusemens. 1749.  x  thProb. 108, p. 249. (Double and give away 6) thrice to leave 0. Prob. 111, pp. 252253. Double and spend 20, triple and spend 27, double and spend 19, leaving 250. Prob. 118, p. 260. (Halve and give 1/2 more) thrice to leave 0.  x  tWalkingame. Tutor's Assistant. 1751.  x  th1777: p. 82, prob. 7; 1860: p. 111, prob. 7. Stealing apples. Give half and get back 10, give half and get back 4, give half, get back 1, leaving 24. 1777: pp. 174175, prob. 86; 1860: p. 183, prob. 85. Sheep fold robbed of half its sheep and half a sheep more, thrice, leaving 20.  x  tEuler. Algebra. 1770. I.IV.III: Questions for practice, no. 9, p.204. (Spend 100, then gain @) thrice to double original value. Vyse. Tutor's Guide. 1771?  x  thProb. 18, 1793: p. 32; 1799: p. 36 & Key p. 29. Sheep fold. (Lose half and  more) thrice, leaving 20. Prob. 31, 1793: p. 57; 1799: p. 62 & Key p. 69. (Gain @, less 100) for 3 years to yield 3179 11s 8d. Solution assumes the final quarter year is the same as (gain1/12, less 25), but it is not obvious how to determine an appropriate expression for the quarterly effect. In general, repeating ax b four times gives a4x b(a4ĩ1)/(a1) and setting this equal to 4x/3 100 gives a = 1.075.., b=22.371. However, the 100 is the expenses of the merchant's family and he may not be able to reduce it in one quarter. Prob. 33, 1793: pp. 5758; 1799: pp. 6263 & Key p. 70. Lose , get back 10; lose@, get back 2; lose , get back 1; leaving 12. Prob. 2, 1793: p. 129; 1799: p. 137 & Key p. 179. (Double and spend 6) thrice, leaving 0. Solution by double position. Prob. 4, 1793: p. 129; 1799: p. 137 & Key p. 180. Lose , gain 10; lose , gain 4; lose , gain 1; yielding 18. Solution by double position.  x  tDodson. Math. Repository. 1775.  x  thP. 10, quest. XXIV. Double and spend 6; triple and spend 12; quadruple and spend 18; leaving 30. P. 47, quest. C. Shepherd loses  of his flock and  of a sheep; then @ of his flock and @ of a sheep; then  of his flock and  of a sheep; and has 25 sheep left. P. 48, quest CI. Man (spends 50 and gains @ on the remainder) thrice, yielding double his original amount. P. 49, quest. CII. Lose , win 3, lose @, win 2, lose 1/7, yielding 12.  x  tOzanam-Montucla. 1778.  x  thProb. 15, part a, 1778: 207208; 1803: 203. Prob. 14, 1814: 175-176; 1840: 91. (Sell half the eggs and half an egg more) thrice to leave 36. This was one of the more popular forms of the puzzle after this time see: Jackson, Endless Amusement II, Nuts to Crack, Young Man's Book, Boy's Own Book, Magician's Own Book, Boy's Own Conjuring Book, Wehman, Collins, Sullivan. Prob. 15, part b, 1778: 208209; 1803: 203204. Prob. 15, 1814: 176177; 1840: 91. (Spend half and 1/2 more) thrice leaving 0. Gives the rule for the problem with more iterations.  x  tBonnycastle. Algebra. 1782. P. 86, no. 20 (1815: p. 107, prob. 30). Lose 1/4 of what he has, win 3, lose 1/3, win 2, lose 1/7, leaving 12. Eadon. Repository. 1794.  x  thP. 296, no. 9. Man loses 1/4, then gains 3, then loses 1/3, then gains 2, then loses 1/7, then has 12. P. 296, no. 10. Man (spends 50 and then gains 1/3) thrice to double his money. P. 297, no. 14. Man sends out 1/3 and 25 more of his men, leaving 1/2 and 100 more.  x  tBonnycastle. Algebra. 10th ed., 1815. P. 106, no. 19. (Double and pay 1) four times, leaving 0. Jackson. Rational Amusement. 1821. Curious Arithmetical Questions.  x  thNo. 10, pp. 16 & 73. (Give one half) seven times, leaving 1. No. 13, pp. 17 & 74. (Give half plus i) for i = 1, 2, 3, leaving 1. No. 18, pp. 18 & 75. (Spend half plus half a guinea) four times, leaving 0. No. 19, pp. 1819 & 7576. (Sell half, receive 10); (sell third, receive 2); (sell half, receive 1); leaving 12. No. 23, pp. 1920 & 77. Same as OzanamMontucla, prob. 15a.  x  tEndless Amusement II. 1826? Prob. 27, pp. 202203. Identical to the 1803 English of OzanamMontucla, prob. 15a. = New Sphinx, c1840, p. 138. Nuts to Crack II (1833), no. 126. Same as OzanamMontucla, prob. 15a. Young Man's Book. 1839. Pp. 234235. Identical to the 1803 English of OzanamMontucla, prob. 15a. Boy's Own Book. 1843 (Paris): 344. Same as OzanamMontucla, prob.15a. = Boy's Treasury, 1844, p. 301. = de Savigny, 1846, p. 289: La paysanne et les ufs. Magician's Own Book. 1857.  x  thCountrywoman and eggs, pp. 238239. Almost identical to Boy's Own Book. The old woman and her eggs, p. 240. (Give half and half an egg more) thrice, leaving 1. The apple woman, p. 252. Sell half, gain 10, sell third, gain 2, sell half, gain 1, have 12 remaining. = Book of 500 Puzzles, 1859, p. 66. = Illustrated Boy's Own Treasury, 1860, prob. 15, pp. 428 & 431432.  x  tBoy's Own Conjuring Book. 1860.  x  thCountrywoman and eggs, p. 205. Almost identical to Boy's Own Book. The old woman and her eggs, p. 212. Identical to Magician's Own Book. The apple woman, p. 223. Identical to Magician's Own Book.  x  tVinot. 1860. Art. LV: Les ufs, pp. 7273. (Sell half the eggs and half an egg more) thrice to leave 0. Lewis Carroll. Letter of 22 Jan 1878 to Jessie Sinclair. = CarrollCollingwood, pp. 205207. Cf CarrollWakeling, prob. 26, pp. 34 & 72. "Tell Sally it's all very well to say she can do the two thieves and the five apples, ...." Wakeling omits the number of apples since it is the answer to the problem he poses. Cohen and Wakeling give a possible version of the problem, provided by Peter Heath, as: steal half and half an apple more, then the second thief steals half of what the first thief stole and half an apple more, leaving none, for which the answer is 5. Cohen says John Fisher [The Magic of Lewis Carroll; op. cit. in 1, p. 79] cites a similar problem from the notebooks of Samuel Taylor Coleridge, but this is: (sell half and half an egg more) thrice, leaving three. Another possibility, which seems much more likely to me, would be (steal half and half an apple more) twice, leaving 5, for which the answer is 23. This is the more common form of the problem, whereas Heath's version takes 5 as the answer rather than the data. In CarrollGardner, pp.77-78, Gardner gives a totally different explanation, saying this is an old magic trick and explaining it. Mittenzwey. 1880.  x  thProb. 103, pp. 21 & 73; 1895?: 120, pp. 2526 & 75; 1917: 120, pp. 24 & 7273. Egg woman sells (half of her eggs and half an egg more) four times, leaving 1. In 1917, the solution is expanded and he notes that starting with 2k 1, four stages bring you to 2k4 1, but he doesn't seem to understand how to solve the problem in general. Prob. 120, pp. 2425 & 76; 1895?: 138, pp. 2829 & 79; 1917: 138, pp. 26 & 76-77. Three men successively taking 1/3 of a pile of potatoes. Remainder is 24. Observes that the second person is entitled to 3/8 of the remainder and the third person gets the rest [but this is unnecessary information]. How many potatoes were there?  x  tWilliam J. Milne. The Inductive Algebra Embracing a Complete Course for Schools and Academies. American Book Company, NY, 1881. Pp. 138 & 332, no. 81. (Double and lose one) thrice to get triple original amount. Hoffmann. 1893. Chap. IV gives several deterministic examples: nos. 27, 39, 46, 67, 76(?see 7.E.1), 111, 112. Lucas. L'Arithm)tique Amusante. 1895. P. 184: Prob. XLII: La marchande d'ufs. (Sell 1/2 plus half an egg more) n times to leave 0. CarrollWakeling. 1888 Prob. 17: Four brothers and a monkey, pp. 21 & 68. This has a pile of nuts on a table and is Form 1, Ending 0. Wakeling gives the solution 765 and says there are other solutions, citing 2813 and 5885, i.e. 765 + 2048 k, but the general solution is actually 765+ 1024 k. This is one of the problems on undated sheets of paper that Carroll sent to Bartholomew Price. Wakeling said he will look for a watermark on it, but the date is now pretty definite. Wakeling says there is no other mention of the problem in Carroll's work, MSS or correspondence. CarrollGardner, p. 53, mentions CarrollWakeling and cites his 1958 article. On 28 May 2003, Wakeling kindly sent me copies of three items of the Carroll/Price material. First is Carroll's solution of the problem, which is typewritten, 'probably using Dodgson's Hammond typewriter, purchased in 1888.' This solution is grossly erroneous he only takes three stages and obtains the answer 61 + 64k. Most importantly, Wakeling sent a note from John (later Sir John) Evans to Price, dated 15 Oct 1888, thanking Price for his solution of the problem and saying that his attempt had gotten to a value of 1789 (which is a correct solution!). Evans then adds that he cannot make Price's solution work. Price must have given 253, but after the fourth brother, there remain 78 which is not divisible by four (nor is it one more than a multiple of four). Evans than says that 509 (= 253 + 256) and 765(=253+ 512) also fail, 'I think'. Wakeling also sent the statement, only, of the problem, in Evans' handwriting, headed Four Brothers & the Family Monkey this differs from the version in CarrollWakeling. Though this is a moderately messy problem, it is depressing to discover that three competent mathematicians were unable to get the correct solution and failed to check the solutions that they had obtained!! However, we now know that the problem was in circulation in 1888, and the fact that wrong answers were being obtained shows that the problem was new at that time. W. W. Rouse Ball. Elementary Algebra. CUP, 1890 [the 2nd ed. of 1897 is apparently identical except for minor changes at the end of the Preface]. Prob. 12, p. 260 & 475. Three Arab jugglers and their monkey, on their way to Mecca, buy a basket of dates. Ending 1. He gives no background to the problem nor any indication that it is novel. The solution is just the numerical answer. The presence of two versions by c1890 would indicate the problem must have been known to at least a few other people. The fact that both Carroll and Ball knew of the problem leads one to conjecture a possible mode of transmission. After the appearance of Alice in 1865, Carroll's reputation was immense. Ball's A Short Account of the History of Mathematics appeared in 1888 and was wellknown in the Englishspeaking world and Ball was becoming known as an authority on the history of mathematics and on mathematical recreations. Either or both might have been sent the problem from India (or anywhere in the world), perhaps by the translator of Mahavira when he first came on the problem. However, I have examined the Ball material at Trinity College Cambridge and it is clear that he (or his heirs) disposed of his correspondence, so this conjecture cannot be verified. We hope 'something will turn up' to elucidate this. Something has turned up see the further material under CarrollWakeling but this does not determine the source of the problem. Clark. Mental Nuts. See also at 1904.  x  th1897, no. 11; 1904, no. 19; 1916, no. 17. The man and his money. (Spend 1/2 and 1/2 more) four times to leave 0. 1897, no. 20; 1904, no. 92; 1916, no. 8. The man and the stores. (Double and pay 10) thrice to leave 0. 1897, no. 29; 1904, no. 40. The farmer and his horses. (Pay 1, then sell 1/2, then pay 1 more) four times to leave 1.  x  tDudeney. Problem 522. Weekly Dispatch (8 & 22 Nov 1903) both p. 10. Multiple of 25 eggs. Sell half and half an egg more until all gone. Clark. Mental Nuts. 1904, no. 99. Three boys and basket of apples. Coconuts Ending 1 with 3 people. (This is not in the 1897 or 1916 eds. This complicates the possible connection with Mahavira cf the discussion under Pearson, below.) Pearson. 1907. Part II. Several determinate versions, and the following.  x  thNo. 29: The men, the monkey, and the mangoes, pp. 119 & 197. Coconuts Ending 1 with 3 people. Gives only one solution. No. 94: One for the parrot, pp. 133-134 & 210. Coconuts Ending 1 with 4 boys, a bag of nuts and a parrot. Gives only one solution. The connection of these with Mahavira, 850, bemused me and I conjectured that Pearson might have heard of Mahavira's work, though the translation didn't appear until 1912. Kaye's note (see under Mahavira in the Abbreviations) shows that an advance version of the translation was produced in 1908, which makes my conjecture much more likely. The Frontispiece of Pearson's book shows him as a clergyman of about 4050 years old, and another of his books describes him as MA of Balliol College, Oxford and Rector of Drayton Parslow, Buckinghamshire, with a stamp underneath giving Springfield Rectory, Chelmsford [Essex], so he might have been a missionary or had Indian contacts. (Incidentally, the publisher Cyril Arthur Pearson was his son.) HOWEVER, I have now seen Carroll, Ball and Clark and this makes the connections less clear.  x  tWehman. New Book of 200 Puzzles. 1908.  x  thP. 50: The sheepfold robbery. (Lose 1/2 and 1/2 a sheep more) thrice leaving 2. P. 51: The maid and her apples. c= Magician's Own Book, p. 252. P. 57: The countrywoman and her eggs. Same as OzanamMontucla, prob.15a.  x  tNelson L. Roray, proposer; A. M. Harding, Norman Anning and the proposer, solvers. Problem 288. SSM 12 (1912) 235 & 520-521. Coconuts Ending 1 with 3 men. Anning shows that the solution is -2 (mod 34), but none of the solvers generalise to n men. Loyd. Newsboys puzzle. Cyclopedia, 1914, pp. 116 & 354. (= MPSL2, prob. 9, pp. 8 & 123. = SLAHP: Family rivalry, pp. 51 & 103.) Complex specification of one amount. Loyd. A study in hams. Cyclopedia, 1914, pp. 268 & 375. (= SLAHP: The Ham peddler, pp.81 & 117.) (Half plus half a ham more) four times, (half a ham plus half), (half plus half a ham), leaving 0. R. L. Weber. A Random Walk in Science. Institute of Physics, London & Bristol, 1973. P.97 excerpts a Russian book on humour in physics which states that P. A. M. Dirac heard a version of the problem with three fishermen and a pile of fish, but only three divisions, at a mathematical congress while he was a student (at Cambridge?) and gave the solution, 2. In fact, he only came to Cambridge as a graduate student in 1923 and became a fellow in 1927, so that the story, if true and if it refers to his time at Cambridge, relates to the mid 1920s. Ben Ames Williams. Coconuts. Saturday Evening Post (9 Oct 1926) 10,11,186,188. Reprinted in: Clifton Fadiman, ed.; The Mathematical Magpie; Simon & Schuster, NY, 1962, pp. 196214. Ending 0 with 5 men. R. S. Underwood, proposer; R. E. Moritz, solver. Problem 3242. AMM 34 (1927) 98 (??NX) & 35 (1928) 47-48. General version of coconuts problem, Ending 0 with n men. Wood. Oddities. 1927.  x  thProb. 52: A goose problem not for geese to solve, pp. 43 & 44. Sell a half and half a goose more; sell a third and a third of a goose more; sell a quarter and 3/4 of a goose more; sell a fifth and a fifth of a goose more; leaving 19. Prob. 57: Eggs this time, p. 46. Sell half and half an egg more; sell a third and a third of an egg more; sell a quarter and a quarter of an egg more; sell a fifth and a fifth of an egg more; leaving a multiple of 13. Determine the least number of possible eggs. Gives answer 719. Complete answer is 719 (mod 780).  x  tCollins. Book of Puzzles. 1927. The basket of eggs puzzle, p. 77. Same as OzanamMontucla, prob. 15a. Collins. Fun with Figures. 1928. The parrot talks, pp. 183185. Four boys and a parrot and a bag of nuts. Ending 1 with n = 4. = Pearson 94. Kraitchik. La Math)matique des Jeux, 1930, op. cit. in 4.A.2. P. 13.  x  thProb. 39. Monkey and mangoes problem, Ending 1 with 3 men. = Pearson 29. = MR, 1942, prob. 35, pp. 3233. Prob. 40. (Take @) thrice, leaving 8. = MR, 1942, prob. 36, p. 32. He asserts these are Hindu problems but gives no source.  x  tRudin. 1936. No. 5, pp. 3 & 76. Three men. (Take away @) thrice, then divide in thirds. He gives only the answer 81, though any multiple of 81 works. Hubert Phillips. Question Time. Op. cit. in 5.U. 1937. Prob. 203: Adventure island, pp. 137 & 246. Ending 1 with 5 men and Friday instead of a monkey. Francis & Vera Meynell. The Week-End Book. Nonesuch Press, 1924 and numerous printings and editions. I have 8th printing, 2nd ed., Mar 1925, and a 5th(?) ed., in 2 vols., Penguin, 1938. The earlier edition has some extra text surrounding the problems, but has only 8 of the 12 problems in the Penguin ed. This problem is not in the 2nd ed. 5th?? ed., prob. seven, p. 408: Three men and a monkey. Ending 0, with 3 men. No solution. Joseph Bowden. Special Topics in Theoretical Arithmetic. Published by the author, Lancaster, Pennsylvania, 1936. The problem of the dishonest men, the monkeys and the coconuts, pp. 203212. ??NYS cited by Pedoe, Shima & Salvatore. McKay. At Home Tonight. 1940.  x  thProb. 33: The niggers and the orchard, pp. 69 & 82. Three men and apples. Ordinary division with the extra thrown away and Ending 1. Prob. 35: Dividing nuts, pp. 70 & 83. Divide nuts among 5 girls with one left over. One girl divides hers among the rest, with one left over. Then another girl divides hers among the rest, with one left over.  x  tThe Home Book of Quizzes, Games and Jokes. Op. cit. in 4.B.1, 1941. Pp.149-150: The bag of peanuts. Five Italians, a bag of peanuts and a monkey. Coconuts Ending 1 with 5 people. The given answer is 3121, but it should be 15621. 3121 would be the answer for Ending 0 cf Leeming, 1946. Garrett Birkhoff & Saunders Mac Lane. A Survey of Modern Algebra. Macmillan, (1941, ??NYS), revised, 1953. Prob. 11, p. 26 (p. 28 in the 4th ed. of 1977). Usual five men and a monkey form, but no ending is specified. That is, nothing is stated about the number left over in the morning except that it is an integer. No answer given, but there is a hint to try 4 coconuts. Somewhat surprisingly, the answer is the same as for the Ending 0 problem see Pedoe, Shima & Salvatore. Leeming. 1946. Chap. 5, prob. 16, pp. 57-58 & 177. Five Italian organ grinders, their monkey and a pile of peanuts. Coconuts Ending 1 with 5 people. The given answer is 3121, but it should be 15621. 3121 would be the answer for Ending 0 cf Home Book, 1941. Sullivan. Unusual. 1947. Prob. 40: Another old problem. Sell (half plus half an egg more) thrice, leaving 36. = OzanamMontucla, prob. 15a. Paul S. Herwitz. The theory of numbers. SA 185:1 (Jul 1951) 5255. With letter and response in SA 185:3 (Sep 1951) 2 & 4. Gives the Birkhoff & Mac Lane version with n= 3 and solves it by explicitly computing the number remaining in terms of the initial number, obtaining one linear diophantine equation in two unknowns. He states the equation for the case m = 4, but doesn't give the solution, and for the general case, though he doesn't sum the geometric progression that appears. The letter requests the solution for the case m = 5 and Herwitz outlines how to find the solution by the Euclidean algorithm, obtaining 3121. Anon.?? Monkeys and coconuts. Mathematics Teacher 54 (Dec 1951) 560562. ??NYS cited by Pedoe, Shima & Salvatore. Anonymous. The problems drive. Eureka 17 (Oct 1954) 89 & 1617. No. 2. Three men and cigarettes guarded by a Boy Scout. Two are given to the scout at each division and at the end, so this is Form 1c, Ending c, with c = 2. Solution observes that (n1)c is a fixed point and the solution is (n1)c (mod nn+1), as seen in Singmaster, but the solution here doesn't give any proof. Ron Edwards. The cocoanut poker deal. The Cardiste (Mar 1958) 56. Uses the three person problem as the basis of a card trick. He states the original problem in a novel form each hunter finds the pile evenly divisible by three, so the monkey doesn't get any coconuts until the morning division when he gets one. But in the trick, Edwards uses the classic form, with a variation. The 52 cards are dealt into 3 piles, with one extra put in a discard pile. The spectator places one of the end piles on the middle giving a new deck of 34. The process is repeated twice more leaving 22, then 14 cards. These 14 are dealt into three piles, but now there are two extras which are discarded and then the five discards are found to be a royal flush! (The Cardiste was a mimeographed magic magazine which ran from 1957 to 1959. My thanks to Max Maven for remembering and finding this and sending a copy.) M. Gardner. SA (Apr 1958) = 2nd Book, chap. 9. Describes the Williams story of 1926 and says the Post received 2000 letters the first week after publication and the editor telegrammed: "For the love of Mike, how many coconuts? Hell popping around here." Gardner says Williams modified the older problem of Ending 1 with 5 men. He gives the solution of 4, but says he could not trace its origin. His addendum cites Anning (1912) for this. Roger B. Kircher. The generalized coconut problem. AMM 67:6 (Jun/Jul 1960) 516519. Generalizes by taking any number of sailors, any number of divisions and allowing the i-th division to discard a variable amount Vi before taking away 1/n of the rest, even allowing negative Vi, e.g. if the monkey is adding coconuts to the pile! Sadly, his basic recurrence equations (1) and (2) are misprinted. He solves this by use of difference calculus. Nathan Altshiller Court. Mathematics in Fun and in Earnest. Op. cit. in 5.B. 1961. Prob. c, pp. 188190. Take @ thrice, leaving 8. How to divide the remainder to make things equal? Philip Haber. Peter Pauper's Puzzles & Posers. Peter Pauper Press, Mount Vernon, NY, 1963. Prob. 125, pp. 34 & 57. Basket of pears divided among four people. First gets  of the total plus  of a pear. Second gets @ and @. Third gets  and . Fourth gets remainder, which is half of what the first got. Harold H. Hart. Grab a Pencil No. 3. Hart Publishing, NY, 1971. The horse trader, pp. 41 & 118. (Pay 1, then half, then 1 more) thrice, leaving 1. Birtwistle. Math. Puzzles & Perplexities. 1971.  x  thThe Arabs, the monkey and the dates, pp. 5052. Four people, Ending 1. Gets 1021. Baling out, pp. 52, 169 & 189. (Lose @ of a load of bales of hay and @ of a bale more) four times, leaving an integral number and no broken bales. Solution is to start with 80 bales.  x  tShakuntala Devi. Puzzles to Puzzle You. Op. cit. in 5.D.1. 1976. Prob. 144: The mango thieves, pp. 89 & 135. Ending 1 with three boys, but no monkey the boys each eat a mango when they steal @ and there is an extra mango when they divide in the morning. Michael Holt. Figure It Out Book One. Dragon (Granada), London, 1978. Problem 14 (no page number) gives a Russian version involving a man who sells his soul to the Devil. (Double and spend 8) thrice to leave 0. Birtwistle. Calculator Puzzle Book. 1978. Prob. 27: Shipwreck, pp. 21 & 80. Four sailors, the ship's cat and a box of biscuits, but only three of the sailors take a fourth, with one for the cat, before the final division into four, with none for the cat. Answer is 61, but this should be taken (mod 256). Dan Pedoe, Timothy Shima & Gali Salvatore. Of coconuts and integrity. CM 4:7 (Aug/Sep 1978) 182-185. General discussion of history and examples, based on Gardner (1958). They generalise to consider giving c to the monkey and to having m divisions. They first do Birkhoff & Mac Lane's version and Ending 1, effectively noting that the latter is the same as the former but with an extra division step. I.e., the Birkhoff & Mac Lane problem has m = n, while Ending 1 has m = n+1. Examination of the results shows that the Birkhoff & Mac Lane problem with odd n has the same answer as the Ending 0 problem, but for even n, it corresponds to an Ending 2 problem, which turns out to be one greater than the Form, Ending = 0, 1 problem. They separately solve the Ending 0 problem, again with general c. Ben Hamilton. Brainteasers and Mindbenders. (1979); Prentice-Hall, Englewood Cliffs, NJ, 1981. Problem for March 29, pp. 36 & 156. i-th customer buys i+1 plus 1/(i+1) of the rest. How many customers can be served? Scot Morris. The Next Book of Omni Games. Plume (New American Library), NY, 1988. The monkey and the coconuts, pp. 3031 & 182183. Sketches usual history. He then notes that the usual process has the pile  1 (mod n), so the monkey essentially gets one, then the pile is divided into n parts. But one could alternatively have the pile  0 (mod n), so the pile is divided into n parts, the sailor takes his part and then the monkey takes 1 from the remainder. I.e. rather than removing one and 1/n of the rest, each step removes 1/n plus one more. I have now termed these Form 1 and Form 0. In this situation he doesn't allow the monkey to get one in the final division, i.e. he considers Ending 0, but Ending 1 could be permitted, as studied by me below. Liz Allen. Brain Sharpeners. Op. cit. in 5.B. 1991. It really takes the biscuit, pp. 5052 & 119. Four boys and their cat dividing biscuits. Form 1, Ending 1 with 4 boys. David Singmaster. Coconuts. The history and solutions of a classic diophantine problem. Technical Report SBUCISM9302, School of Computing, Information Systems and Mathematics, South Bank Univ., Dec 1993, 21pp, Feb 1994. Submitted to Mathematics Review (Univ. of Warwick), 1993, but the journal closed before using it. Revised in 1996 and again as the 3rd ed. of 11 Sep 1997, 21pp. Extensive history, based on the material in this section. Following Morris's comment, I consider both division Forms and both Endings, giving four basic problems rather than the three that he mentions. In the 3rd ed., I changed the terminology to Form and Ending and have converted the material in this section to conform with this. A little reflection shows that the solution for Form 0, Ending 0 is one less than for Form 1, Ending1. Actual calculation shows that one of the four cases has the same sequence of pile sizes as another, but shifted by one stage. When n is odd, Form 0, Ending 0 is the same as Form 0, Ending 1, but starting one stage earlier. When n is even, Form 1, Ending 1 is the same as Form 1, Ending 0, but starting one stage earlier. Although these results are easily seen from the algebraic expressions, I cannot see any intuitive reason for these last equalities. Having now seen Pedoe, Shima & Salvatore, I have added two supplementary pages discussing the Birkhoff & Mac Lane problem and relating it to the standard versions. 7.E.1.VERSIONS WITH ALL GETTING THE SAME  tThe i-th child gets some linear function of i applied to the remainder, but all wind up with the same amount. See Tropfke 586.  tFibonacci. 1202.  x  thP. 279 (S: 399). ith gets i + 1/7 of rest. (Sanford 219 gives the English; H&S 61-62 gives Latin & English.) P. 279 (S: 399). ith gets i + 2/11 of the rest, but he doesn't ask for the number of children. Pp. 279-280 (S: 399401). ith gets (3i-1) + 6/31 of the rest. Pp. 280-281 (S: 401). ith gets (2i+1) + 5/19 of the rest.  x  tMaximus Planudes. 0--# )'  +%       [Psephephoria kat' Indous e Legomene Megale (Arithmetic after the Indian method)]. c1300. (Greek ed. by Gerhardt, Das Rechenbuch des Maximus Planudes, Halle, 1865, ??NYS [Allard, below, pp. 2022, says this is not very good]. German trans. by H. Waeschke, Halle, 1878, ??NYS [See HGM II 549; not mentioned by Allard].) Greek ed., with French translation by A. Allard; Maxime Planude Le Grand Calcul selon des Indiens; Travaux de la Facult) de Philosophie et Lettres de l'Univ. Cath. de Louvain XXVII, Louvain-la-Neuve, 1981. ith gets i + 1/7 of the rest, pp. 191-194 & 233-234. On pp. 233-234, Allard discusses the history of the problem, citing: Fibonacci; BR; Parisinus supp. gr. 387 & Scoriolensis , I 16. BR. c1305. No. 84, pp. 102-105. ith gets i + 1/7 of the rest. Gherardi. Libro di ragioni. 1328. Pp. 37-38. ith daughter gets i + 1/10 of the rest. Lucca 1754. c1330. F. 82v, p. 199. i-th gets i + 1/10 of rest. (This problem is not clearly expressed.) Bartoli. Memoriale. c1420. Prob. 9, f. 75v (= Sesiano pp. 138 & 147148). ith gets i + 1/7 of the rest. Pseudodell'Abbaco. c1440.  x  thProb. 168, p. 140. ith gets 1000 i + 1/10 of the rest. Prob. 169, pp. 140-141. ith gets 1/6 plus 10 i.  x  tAR. c1450. No. 114, 115, 352, pp. 64-65, 154, 173-174 & 220.  x  th114: ith gets i + 1/10 of rest. 115: ith gets i + 1/6 of rest. 352: ith gets 1/5 of remainder + i - 1.  x  tMuscarello. 1478. F. 85v, pp. 204205. ith gets i + 1/9 of the rest. Chuquet. 1484. Probs. 129-141. English of prob. 129 in FHM 224225, with some description of the others. ith child gets (ai + b) plus r of the rest; ith child gets r of amount and a + bi more. Many problems have non-integral number of children and amounts received e.g. prob. 133 has 2 5/6 children receiving 6 2/3, with the 5/6 getting 5 5/9. HB.XI.22. 1488. Pp. 44-45 (= Rath 247). ith gets i + 1/9 of rest. Calandri. Arimethrica. 1491. F. 65r. ith gets 1/10 + 1000 i Calandri, Raccolta. c1495. Prob. 26, pp. 25-26. ith gets 1000 i + 1/10 of the rest. Ghaligai. Practica D'Arithmetica. 1521.  x  thProb. 24, ff. 65v66r. ith gets 1000 i + 1/7 of rest. Prob. 25, f. 66r. ith gets 1/7 + 1000 i.  x  tCardan. Practica Arithmetice. 1539. Chap. 66, section 65, f. FF.ii.r (p.155). ith child gets 1/7 of remainder + 100 i Tartaglia. Quesiti, et Inventioni Diverse. Venice, 1546. Book 9, quest. 2, pp. 98r-98v. ith child gets i + 1/8 of the rest. Tartaglia. General Trattato, 1556, art. 46, pp 245v-246r. ith child gets i + 1/6 of the rest; discusses same with 1/7 and 1/13. Buteo. Logistica. 1559. Prob. 78, pp. 286288. ith child gets 1/6 + 100 i. Bachet. Problemes. 1612. Addl. prob. VII: Un homme venant ! mourir ..., 1612: 149154; 1624:221226; 1884: 158-161. ith child gets i + 1/7 of the rest; also aiĠ + 1/n and 1/n + ai. Asserts some cases are impossible, contrary to Chuquet's approach. Labosne has much revised the entire problem. Ozanam. 1725. Prob. 10, question 9, 1725: 67-68. Prob. 1, 1778: 185; 1803: 182183; 1814: 159; 1840: 82. ith gets 10000 i + 1/7 of the remainder. Les Amusemens. 1749.  x  thProb. 55, pp. 187188. ith gets 1000 i + 1/7 of the rest. Prob. 177, p. 328. ith gets 1000 i + 1/5 of the rest.  x  tEuler. Vollstndige Anleitung zur Algebra. (St. Petersburg, 1770) Part 2, sect. 1, chap. 3, art.42. (= Opera Omnia (1) 1 (1911), pp. 226-228. = Algebra; 1770; I.IV.III.604: Question 21, pp. 202-203.) ith gets 100 i + 1/10 of rest. Hutton. A Course of Mathematics. 1798? Prob. 10, 1833: 214217; 1857: 218221. Father with three sons leaves ai + 1/n of the remainder to the ith and this exhausts the fortune (but they do not get equal amounts except when n = 4). Finds algebraic expressions for the total and each portion, e.g. the total fortune is (6n2 4n + 1)a/(n 1)2. Augustus De Morgan. Arithmetic and Algebra. (1831 ?). Reprinted as the second, separately paged, part of: Library of Useful Knowledge Mathematics I, Baldwin & Craddock, London, 1836. Art. 114, pp. 2930. First gets 2 + 1/6 of rest; second gets 3 + 1/6 of rest; they find they got the same amount. Bourdon. Alg/bre. 7th ed., 1834. Art. 48, pp.6671 is the same as Hutton's 1798? problem. Art. 49, pp. 7173 then treats the usual problem of the same form, finding n1 children and a fortune of a(n1)2. Fred Barlow. Mental Prodigies. Hutchinson, London, (1951), corrected reprint, 1952. On pp.3841, he describes Henri Mondeaux (18261862), an illiterate who developed powers of mental calculation, and was then taught by a schoolmaster, M. Jacoby. At some time, he was asked to solve the problem of people taking 100i + 1/13 of the rest and he found the answer in a few seconds by taking 12 x 100 as the amount of each person and 12 x 12 x 100 as the total amount. I wonder if he knew this type of problem beforehand?? Vinot. 1860. Art. XXXIX: Du testament, pp. 3839. ith gets 1000 i + 1/5 of the rest. Hoffmann. 1893. Chap. IV, no. 76: Another eccentric testator, pp. 166 & 221-222 =HoffmannHordern, p. 148. First son gets 1/6 plus 240, second son gets 1/5 of the remainder plus 288, ..., fifth gets 1/2 of the remainder plus 720 and all wind up with the same amount. Lucas. L'Arithm)tique Amusante. 1895. Prob. XL: Le Testament du nabab, pp. 144147. i-th gets i + 1/7 of the rest. Gives general solution of: ith gets i + 1/n of the rest, using a rectangular layout of markers similar to ancient Indian multiplication tables and Lucas thinks the ancient Indians must have known this problem and solution. 7.F. ILLEGAL OPERATIONS GIVING CORRECT RESULT 'Two digit' refers to illegal cancellation with two digit numbers, e.g. 16/64 = 1/4, etc. A. Witting. Ernst und Scherz im Gebiete der Zahlen. Zeitschr. math. u. naturw. Unterricht 41 (1910) 45-50. P. 49 gives the rule of Ahrens, below, pp. 75-76, for the case k = 2. He also gives three of the two digit examples: 26/65, 16/64, 19/95 omitting 49/98. Ahrens. A&N, 1918.  x  thPp. 73-74 finds the two digit solutions and some with more digits. Pp. 75-76 studies (a + m/n)1/k = a * (m/n)1/k and finds that m= a, n = ak - 1 works.  x  tW. Lietzmann. Lustiges und MerkwGrdiges von Zahlen und Formen. 1922. 2nd ed., F. Hirt, Breslau, 1923. Pp. 103-104. Gives 26/65 and 16/64 and the general rule of Ahrens, pp. 75-76, citing Witting and Ahrens. 4th ed, same publisher, 1930, p. 153, says there are more two digit examples and gives also 266/665, etc. The material is also in the 6th ed. (1943), but not in the 7th ed. (1950). R. K. Morley, proposer; Pincus Schub, solver. Problem E24. AMM 40 (1933) 111 & 425-426. 2 digit versions. G. [presumably the editor, Jekuthiel Ginsburg]. Curiosa 31 Another illegal operation. SM 5 (1939) 176. Cites Morley. Refers to E. Nannei presenting several larger examples in 1935, some involving cancellation of several digits. William R. Ransom. Op. cit. in 6.M. 1955. Freak cancellations, pp. 100-102. Finds the 2-digit versions and give examples of several 3-digit forms: 138/345 = 18/45, 163/326= 1/2, 201/603 = 21/63. B. L. Schwartz, proposer; C. W. Trigg, solver. Problem 434 Illegal cancellation. MM 34 (1961) ??NYS & 367-368. 3 digit versions. Anon. Curiosa 122 A common illegal operation. SM 12 (1946) 111. (a2Ġ-b2)/(a-b)=a+b. Alan Wayne, proposer; solution ??NYS. Problem 3568. SSM 75:2 (No. 660) (Feb 1975) 204. (3/2)2 (1/2)2 gives the right answer when the exponents are interpreted as multipliers. Ben Hamilton. Op. cit. in 7.E, 1979. Problem for April 9, pp. 40 & 157-158. 249/498 gives 24/48 correctly, which gives 2/8 wrongly. R. P. Boas. Anomalous cancellation. In: R. Honsberger, ed.; Mathematical Plums; MAA, 1979. Chap. 6, pp. 113-129. Surveys the problem and studies the two digit case for other bases, e.g. 32/13 = 2/1 in base 4. Cites the SM report, 1939. R. P. Boas. Generalizations of the 64/16 problem. JRM 12 (1979-80) 116-118. Summarises the above paper and poses problems. 7.G. INHERITANCE PROBLEMS 7.G.1.HALF + THIRD + NINTH, ETC.  t This is usually called 'The 17 camels', etc. Early versions of this problem simply divided the amount proportionally to the given numbers, regardless of whether the numbers added to one or not. I mention a few early examples of this below. By about the 15C, people began objecting to such proportions, though Tartaglia (qv) and others see no difficulty with the older idea. Sanford 218-219 says Tartaglia was among the first to suggest the 18th camel but I have not found this in Tartaglia so far. H&S 87 says that the use of the 18th camel is really a modern problem. I haven't found it occurring until late 19C, when several authors claim it is centuries old and comes from the Arabic world or India or China! See 1872 below. Not everyone is happy with the problem, even to this day see Ashley, 1997 and many strange explanations have been given. t 1/2, 1/3Pseudodell'Abbaco; Tagliente; Buteo 1/3, 1/4Jackson 9, 8, 7Papyrus of Akhmim 5/6, 7/12, 9/20Recorde 4/5, 3/4, 2/3Tartaglia 44 2/3, 1/6, 1/8Eperson 2/3, 1/2, 1/4Chuquet; Apianus; 1/2, 1/3, 1/4Bakhshali MS; Lucca 1754; AR 204; Calandri; Tagliente; Riese; Recorde; Tartaglia 42; Buteo; Ozanam; Les Amusemens; Decremps; Bullen; Collins; Always; 1/2, 1/3, 1/6AR, 286 1/2, 1/3, 1/9AR, 170, 286; Hanky Panky; Cassell's; Proctor; Cole; Lemon; Hoffmann; Brandreth Puzzle Book; Loyd; H.D. Northrop; Benson; White; Ball-FitzPatrick; Dudeney; Kraitchik; McKay; Sullivan; Doubleday 1; Ashley; 1/2, 1/4, 1/5Lemon; Clark; Ernst; King; Foulsham 1/2, 1/4, 1/6Clark; 1/2, 1/4, 1/8Bath 2/5, 1/3, 1/4AR, 202; Wagner 1/3, 1/4, 1/5BR; Riese; Tartaglia 43; Jackson 1/4, 1/5, 1/6Apianus; W. Leybourn 2/3, 1/2, 1/3, 1/4Papyrus Rhind; Pike; D. Adams, 1835 1/2, 1/3, 1/4, 1/5Chaturveda; Blasius 1/2, 1/3, 1/4, 1/6Mahavira 1/2, 1/3, 1/6, 1/19Parlour Pastime 1/3, 1/4, 1/5, 1/6Walton; Simpson; Dodson; J. King 1/3, 1/4, 1/6, 1/8D. Adams, 1801; 6, 5, 4, 3, 2Mahavira 7/2, 5/2, 15/4, 25/4, 4Papyrus of Akhmim 1/2, 1/3, 1/4, 1/5, 1/6Tonstall 1/3, 1/4, 1/5, 1/6, 1/7Walkingame; Vyse 1/3, 1/4, 1/6, 1/8, 1/9Meyer; HaldemanJulius; Leeming  tPapyrus Rhind, c-1650, loc. cit. in 7.C. Problem 63, p. 101 of vol. 1 (1927) (= p. 53 (1978)). Divide 700 loaves in proportion A::@ : . Papyrus of Akhmim. c7C. Jules Baillet, ed. Le Papyrus Math)matique d'Akhm3m. M)moires publi)s par les membres de la Mission Arch)ologique Fran'ais au Caire, vol. IX, part 1, (1892) 1-89. Brief discussion of this type of problem on p. 56. Probs. 3, 4, 10, 11, 28?, 47, 48, 49 are of this type. I describe two examples.  x  thProb. 3, pp. 6465. Divide 1000 in proportion 3+1/2:2+1/2:3+1/2+1/4:6+1/4 : 4. Prob. 11, pp. 6869. Divide 3 + 1/2 + 1/4 in proportion 7 : 8 : 9.  x  tBakhshali MS. c7C. See in 7.E, where a king gives away  + @ +  of his money! Mahavira. 850. Chap. VI, v. 80, 86, pp. 110111. Divide 120 in proportion 1/2:1/3:1/4:1/6. Divide 480 in proportion 2 : 3 : 4 : 5 : 6. BR. c1305. No. 71, pp. 94-95. 1/3 + 1/4 + 1/5. Lucca 1754. c1330. F. 61r, p. 140. Divide into  + @ + . He divides in proportion 6:4:3. Pseudodell'Abbaco. c1440. Prob. 122, pp. 97-98. Divide into  + @. He divides in the ratio 3 : 2. AR. c1450. Probs. 170, 202204, 207, 229230, 286. Pp. 81-82, 94-96, 106107, 130, 160-161, 166-167, 211-213.  x  th170: 1/2 + 1/3 + 1/9. 202: 1/3 + 1/4 + 2/5. 203: Divide 384 into 2/3 and 6 more, 3/5 and 8 more, 5/6 and 10 more, 7/8 and 6 more. He takes a common denominator of 360 and finds 2/3 of it is 240 and then adds 6 to get 246. Similarly, he gets 224, 310, 321 and then divides in the proportion 246 : 224 : 310 : 321. This is actually indeterminate as it depends on the choice of common denominator. Vogel says the problem is unclear and the solution is false and notes that dividing 387 instead of 384 would give an integral solution. He cites a number of other occurrences of this problem cf. Widman below. 204:  + @ + . 207: Divide 100 into (1/3 - 1/4) + (1/4 - 1/5) + (1/5 - 1/6). 229: Divide 20 into 1, 2, 1, 1, 1, 1. Does as 3:5:2:2:2 : 2. 230: Divide 20 into 1 + @, 2 + , 1, 1, 1, 1. Does as 22:33: 12 : 12 : 12 : 12. 286 discusses problems where one removes fractions and deals with the remainder. Notes that 1/2 + 1/3 + 1/6 leaves nothing, but 1/2 + 1/3 + 1/9 leaves something.  x  tChuquet. 1484. Triparty, part 1. English in FHM 75. "I wish to divide 100 into three parts of such proportion as are 1/2, 2/3, 1/4 ..." Ulrich Wagner. Untitled text known as "Das Bamberger Rechenbuch". Heinrich Petzensteiner, Babenberg (= Bamburg), 1483. Reproduced, with transcription and notes by Eberhard Schr?der as: Das Bamberger Rechenbuch von 1483. Akademie-Verlag, Berlin, DDR, 1988.  x  thPp. 69-70 & 200. = AR, no. 202. Pp. 7172 & 201202. = AR, no. 203.  x  tCalandri. Aritmetica. c1485. Ff. 93v94r, pp. 187-188. Same as Lucca 1754. Johann Widman. Behode und hubsche Rechnung auff allen kauffmanschafft. Conrad Kacheloffen, Leipzig, 1489. ??NYS. (Rara 36-40. This is extensively described by: J.W. L. Glaisher in Messenger of Mathematics 51 (1921-22) 1-148, but he gives the title as: Behode und hubsche Rechenung ....) Smith and Glaisher give Widman, but Knobloch (7.L.2.c) uses Widmann and Behende und hubsche Rechenung ....  x  thF. 195v (Glaisher 1415 & 122). = AR, no. 230. Ff. 195v196 (Glaisher 15 & 122). = AR, no. 207. F. 196v (Glaisher 1819, 38, 45, 122, 130). = AR, no. 203. Glaisher notes that Pacioli's Summa, (see below), gives a more natural determinate interpretation for similar problems. In this example, this would first subtract 6 + 8 + 10 + 6 from 384, leaving 354 which would be divided in the proportion 2/3: 3/5 : 5/6 : 7/8. He also notes the appearance of Widman's problem and solution in Huswirt (1501) ??NYS and of problems similar to Widman and done in the same way, in Arithmetice Lilium (a book of c1510, ??NYS) (divide 100 into 1/2 + 3, 1/3 + 2, 1/5 + 4) and Tonstall. Rudolff's Kunstliche Rechnung of 1526, ??NYS, does (divide into 1/2 and 6, 1/3 and 4, 1/4 less 2) in Pacioli's manner. Cf Apianus for a similar version. Riese's Rechenung nach der Lenge (1550?, ??NYR) does (divide 124 into 2/3 less 12, 1/4 and 10, 5/6 less 24, 3/8 and 6, 2/5 less 7) in Pacioli's way.  x  tPacioli. Summa. 1494. These give the more natural interpretation of this type of problem.  x  thF. 150r, prob. 3. Divide 100 as 1/2 plus 5; 1/3 less 4. Subtracts 1 from 100 and divides the resulting 99 in the proportion 3 : 2. Ff. 150r150v, prob. 4. Divide 100 as 1/2 plus 3; 1/3 less 5. Divides 102 as 3 : 2. F. 150v, prob. 5. Divide 100 as 1/2 less 4; 1/3 less 2. Divides 106 as 3 : 2. F. 150v, prob. 6. Divide 30 as 1/2 plus 2; 1/3 plus 3. Divides 25 as 3 : 2. F. 150v, prob. 7. Divide 10 as 1/2 less 3; 1/3 plus 4. Divides 9 as 3 : 2. F. 150v, prob. 8. Divide 1046 as 1/2 less 2; 1/3 less 1; 1/4 plus 5. No working or answer.  x  tBlasius. 1513. F. F.ii.r: Prime regula. Man leaves 6000 to be divided 1/2 + 1/3 + 1/4 + 1/5. There is an error in the calculation. Tagliente. Libro de Abaco. (1515). 1541.  x  thProb. 94, part 2, ff. 48v49r. Divide 120 into  + @. Prob. 95, ff. 48v49v. Divide 12 into  + @ + .  x  tRiese. Rechnung. 1522.  x  th1544 ed. pp. 81-82; 1574 ed. pp. 55r-55v. 1/3 + 1/4 + 1/5. 1544 ed. pp. 98-99; 1574 ed. p. 66r. Three men take 1/2+1/3+ 1/4 of the profits, making 50 all told. What was the profit? Answer: 50 x 12/13.  x  tTonstall. De Arte Supputandi. 1522.  x  thQuest. 17, pp. 147149. Divide into 1/2 + 1/3 + 1/4 + 1/5 + 1/6. Uses 4350 as common denominator. Quest. 18, pp. 149150. Same as Quest. 17, done with denominator 60. Quest. 21, pp. 151152. Divide into parts: 1/3 + 1/4; 1/4 + 1/5; 1/5 + 1/6. Quest. 22, pp. 153154. Divide 600 into: 2/3 plus 9; 3/5plus 8; 5/6 plus 7; 7/8plus6. See: AR; Widman; Pacioli for discussion of this type of problem. Takes common denominator of 120 and then divides as 89:80:107 : 111 which is not the way I read the problem. I would divide 570 as 80:72:100:105, as done by Pacioli.  x  tApianus. Kauffmanss Rechnung. 1527.  x  thF. H.v.r. Divide 1300 as 1/2 plus 8, 1/3 less 5, 1/4 less 12. Does in Pacioli's manner, dividing 1309 in the proportion 12 : 8 : 6 and then amending by +8, 5, 12. F. H.v.r. Divide 58 as 1/2 + 2/3 + 1/4. Cf Chuquet. F. H.vii.r. Divide 40 as 1/4 + 1/5 + 1/6.  x  tCardan. Practica Arithmetice. 1539. Chap. 66, section 91, second part, f. HH.i.v (p. 166). First has @ plus 7; second has  plus 13; third has minus 28. How much was there? Recorde. Second Part. 1552. Pages from 1668 ed.  x  thPp. 284289: A question of building; An impossible question; The former question of building now possible. Divide 3000 into: 1/2 plus 6; 1/3 plus 12; 2/3less8; 1/4 plus 20. He first says it is impossible, then rephrases it and solves by Pacioli's method. P. 293: Another question of a testament. Divide 7851 into  + @ + . Usual solution. P. 294: Another like question. Divide 450 into 1/2 + 1/3, 1/3 + 1/4, 1/4 + 1/5. Usual solution.  x  tTartaglia. General Trattato. 1556. Book 12, art. 4244, pp. 200r201r.  x  thArt. 42.  + @ + . Long discussion of an error of Luca Pacioli and others who assert that such problems are impossible or illegal. Tartaglia says simply to divide in the proportion 6: 4 : 3 and can't understand why others are making such a fuss. Art. 43. 1/3 + 1/4 + 1/5. Art. 44. 4/5 + 3/4 + 2/3. (Sanford 218-219 says Tartaglia was among the first to suggest the 18th camel but I see nothing of this here. H&S 87 says that this is really a modern problem in that previously property was divided in proportion to fractions, regardless of whether they summed to unity. Tartaglia's discussion of Pacioli and others makes it clear that people were starting to object to this at this time, but examples continue and I don't see the modern version occurring until late 19C.)  x  tButeo. Logistica. 1559.  x  thProb. 5, pp. 203204. Divide 77 into  + @ + . Prob. 74, pp. 283284. Divide 30 into  + @. Discusses the solution. Prob. 75, pp. 284285. Divide 15 as 1/2 plus 2; 1/4 + 3. He divides 10 into  + @, as in Pacioli. See: AR; Widman; Pacioli for discussion of this type of problem. Prob. 76, pp. 285286. Divide 24 as 1/3 less 7; 1/4 less 4. Prob. 77, p. 285. Divide 12 as 2/3 less 3; 1/6 plus 4. Prob. 22, pp. 350351. Divide 60 as 1/4; 1/3 plus 4; 3/4 less 8. Prob. 26, pp. 353354. Divide 30 as 1/2 plus 2; 1/3 less 3. Prob. 28, p. 355. Divide 224 as 1; 6/5 plus 4.  x  tIzaak Walton. The Compleat Angler. (R. Marriott, London, 1653); Everyman edition, Dent, London, 1906, et seq. Chap. V The Fourth Day, pp. 101-102. The World's Classics, OUP, 1935, Chap. V, pp. 114116. Divide 20 into 1/3+1/4 + 1/5 + 1/6. Leaves one left over. W. Leybourn. Pleasure with Profit. 1694. Prob. 11, pp. 3738. 6000 divided 1/4+1/5+1/6. He divides in the proportion 15 : 12 : 10. Cf Apianus. Ozanam. 1725. Prob. 24, question 5, 1725: 179. Divide 26000 into +@ + . Takes in proportion 12 : 8 : 6. Simpson. Algebra. 1745. Section XI (misprinted IX in 1790), prob. XIV, pp. 8586 (1790:prob. XXIV, pp. 8687). Divide 20 into 1/3+ 1/4 + 1/5 + 1/6, done by proportion. Les Amusemens. 1749. Prob. 52, p. 184. Divide 78 into +@ + , by using proportions. Walkingame. Tutor's Assistant. 1751. 1777: p. 177, prob. 119; 1835: p. 180, prob. 58; not in 1860. Divide 500 into 1/3+1/4+1/5+ 1/6 + 1/7. Gives exact answer as integer plus a fraction. 1835 reduces the fractions to lowest terms. Vyse. Tutor's Guide. 1771? 1793: p. 105; 1799: Prob. 6, p. 113 & key p. 151. Same as Walkingame. Solution gives answers rounded to farthings and never gives the exact fractions. Dodson. Math. Repository. 1775. P. 32, quest. LXXVIII. Divide 20s in proportion: 1/3,1/4,1/5, 1/6. Pike. Arithmetic. 1788.  x  thP. 335, no. 4. "Being a little dipped, they agreed that A should pay 2/3, B 1/2, C 1/3, and D 1/4." = D. Adams, 1835. Cf D. Adams, 1801. P. 355, no. 39. A, B, C do a job. A and B do 3/11 of it, A and C do 5/13, B and C do 4/14. (Also entered at 7.H.)  x  tHenri Decremps. Codicile de J)r=me Sharp, .... Op. cit. in 4.A.1. 1788. Avantpropos, pp.1819 mentions +@+, but there is no solution. Samuel Bullen. A New Compendium of Arithmetic ... Printed for the author, London, 1789. Chap. 38, prob. 3, p. 239. Divide into +@ + , phrased as 2A = 3B = 4C. John King, ed. John King 1795 Arithmetical Book. Published by the editor, who is the greatgreatgrandson of the 1795 writer, Twickenham, 1995. The 1795 has dividing 100 into 1/3+1/4+1/5+ 1/6. Gives usual solution. D. Adams. Scholar's Arithmetic. 1801. P. 206, no. 35. Four men divide a purse of $12 as 1/3 + 1/4 + 1/6 + 1/8. Divides in the proportion: 8 : 6 : 4 : 3. Cf his 1835 book. Jackson. Rational Amusement. 1821. Curious Arithmetical Questions, no. 16, pp. 18 & 75. Division in the proportion of 1/3, 1/4, and 1/5, but last person dies. Solution indicates this as standard practice. D. Adams. New Arithmetic. 1835. P. 249, no. 132. = Pike, no. 4. Divides in the proportion: 8 : 6 : 4 : 3. which is the same proportion as in his 1801 version. Parlour Pastime, 1857. = Indoor & Outdoor, c1859, Part 1. = Parlour Pastimes, 1868. Arithmetical puzzles, no. 4, p. 173 (1868: 184). Pay 20s with only 19s by dividing into 1/2 + 1/3 + 1/6 + 1/19. "This, however, is only a payment upon paper." Hanky Panky. 1872. A Chinese puzzle, pp. 7374. 17 elephants to be divided 1/2+1/3+1/9. Cassell's. 1881. P. 102: The clever lawyer. = Manson, 1911, p. 255. 17 horses divided 1/2+1/3 + 1/9. = Rohrbough; Puzzle Craft; 1932, p. 7. Richard A. Proctor. Some puzzles. Knowledge 9 (Aug 1886) 305306. "... the familiar puzzle [of] the farmer, ignorant of numbers, who left 17 horses to his three sons (or, equally well it may be, an Arab sheik who left 17 camels)". Points out that if there were 35 camels, then the Cadi could also be left a camel. E. W. Cole. Cole's Fun Doctor. The Funniest Book in the World. Routledge, London & E.W. Cole, Melbourne, nd [HPL gives 1886 and lists the author as Arthur C. Cole]. P.224: A Chinese puzzle. 17 elephants left by a Chinaman to be divided 1/2+1/3+1/9. Says it is in the Galaxy for August, which might have been an Australian publication by Proctor, who had connections there. Lemon. 1890. The legacy, no. 652, pp. 81 & 121. 19 camels divided 1/2+ 1/4 + 1/5. Don Lemon. Everybody's Pocket Cyclopedia. Revised 8th ed., 1890. Op. cit. in 5.A. P. 135, no. 2. 17 horses: 1/2 + 1/3 + 1/9. Dervish loans them his horse. Hoffmann. 1893. Chap. IV, no. 11: An unmanageable legacy, pp. 147 & 191-192 =HoffmannHordern, p. 119. 1/2+1/3+ 1/9. Answer says "this expedient is frequently employed" in "the Mahomedan Law of Inheritance". Mittenzwey. 1895?. Prob. 164, pp. 34 & 82; 1917: 164, pp. 3132 & 80. 17 camels, 1/2+1/3+ 1/9, dervish loans them his camel. Brandreth Puzzle Book. Brandreth's Pills (The Porous Plaster Co., NY), nd [1895]. P. 3: An unmanageable legacy, with nice colour picture. Identical to Hoffmann. No solution. Loyd. Problem 37: A queer legacy. Tit-Bits 32 (5 Jun & 3 Jul 1897) 173 & 258. =Cyclopedia, 1914, The herd of camels, pp. 57 & 346. 17 horses in proportion 1/2:1/3 : 1/9. Says the use of proportion makes the solution actually correct. Clark. Mental Nuts. 1897, no. 25; 1904, no. 5; 1916, no. 18. The heirs and the sheep. 1897 has 1/2 + 1/4 + 1/6 of 15 sheep. This seems to have been a miscopying of the question with 11 sheep. He says to borrow a sheep and distribute 8, 4, 3, returning one. 1904 & 1916 amend this to 1/2 + 1/4 + 1/5 of 19 sheep. H. D. Northrop. Popular Pastimes. 1901. No. 18: The clever lawyer, pp. 69 & 74. = Cassell's. Benson. 1904. The lawyer's puzzle, p. 225. 1/2 + 1/3 + 1/9. There originally were 18 horses, but one died. William F. White. Op. cit. in 5.E. 1908. Puzzle of the camels, p. 193. 17 camels divided 1/2+ 1/3 + 1/9. BallFitzPatrick. 2nd ed., 1908-1909. Part 1, p. 111, footnote says the problem is Arabic. The material is not in the 1st ed., nor in Ball, 5th ed. A&N, pp. 84-85, cites this but says it has been in German oral tradition for a long time. He gives it with 17 horses. E. Ernst. Mathematische Unterhaltungen und Spielereien. Vol. 2, Otto Maier, Ravensburg, 1912. P. 15: Das geschente Weinfass. Divide 19 in 1/2 + 1/4 + 1/5. Dudeney. MP. 1926. Prob. 89: The seventeen horses, pp. 3334 & 123124. = 536, prob.172, pp. 54-55 & 266-267. Discusses interpretation of proportion, as in Loyd, in detail. King. Best 100. 1927. No. 21, pp. 14 & 43. 19 horses into 1/2+1/4+ 1/5. Collins. Book of Puzzles. 1927.  x  thThe lady bookmaker's problem, pp. 7273. Because 1/2+1/3 + 1/4 = 13/12, one can offer odds in a three horse race of: even money, 2 to 1 and 3 to 1. The sheik and his camels, pp. 7778. Usual form. Cadi loans them his camel.  x  tM. Kraitchik. La Math)matique des Jeux, 1930, op. cit. in 4.A.2, chap.1, prob. 47, p. 15. 17sheep into 1/2 + 1/3 + 1/9. Says it is of Hindu origin. Anon. Foulsham's New Fun Book. W. Foulsham, London, nd [1930s?]. Pp. 85-86: The farmer's horses. Identical to King, 1927. The Bile Beans Puzzle Book. 1933. No. 26: A farmer's will. 19 horses divided 1/2+1/4+ 1/5. McKay. Party Night. 1940. No. 35, p. 184. "It is said that an Arab had 17 cattle." Sullivan. Unusual. 1943. Prob. 12: Will trouble. 17 horses into 1/2+ 1/3 + 1/9. Jerome S. Meyer. Fun for the Family. (Greenberg Publishers, 1937); Permabooks, NY, 1959. No. 30: Think cow it is done, pp. 4243 & 241. Herd to be divided 1/3+1/4+1/6+1/8 + 1/9. Neighbour loans two cows and everything divides up properly with two cows left over for the neighbour. How many cows were there? HaldemanJulius. 1937. No. 131: Cow problem, pp. 15 & 27. Same as Meyer, asking what is wrong with the problem and answering that "The problem is coocoo because all the fractions do not add up to unity." Leeming. 1946. Chap. 5, prob. 24: The herd of cattle, pp. 61 & 179-180. Same as Meyer. Philip E. Bath. Fun with Figures. Op. cit. in 5.C. 1959. No. 66: The vicar's garden, pp. 25 & 52. 7s divided 1/2 + 1/4 + 1/8 by adding an extra shilling. Solution doesn't seem to understand this and claims there really should be 7/8 s left over. Doubleday 1. 1969. Prob. 25: Milk shake, pp. 36 & 159. = Doubleday 5, pp. 3536. 17 cows divided 1/2 + 1/3 + 1/9. He states the usual solution and then asks what is wrong with it. His solution notes that the fractions add to 17/18 and then says 'So, in making his will, the farmer hadn't distributed his entire herd.' This seems confused to me as the entire herd has been distributed to the sons. Jonathan Always. Puzzles for Puzzlers. Tandem, London, 1971. Prob. 35: An odd bequest, pp. 22 & 71. Divide 13 camels as 1/2+1/3 + 1/4. Executor borrows one camel, so there are now 12 camels, then he gives the first son 6 = 12/2 and the second son 4=12/3, leaving 2. But the third son ought to have 3 = 12/4, so the executor returns the borrowed camel and gives the third son his three! The answer says this is an amusing way of arriving at the intended division in the proportion 6 : 4 : 3. See Singmaster, 2000, for an extension. D. B. Eperson. Puzzles, pastimes and problems. MiS 3:6 (Nov 1974) 12-13 & 26-27. Prob.6: The Shah's Rolls-Royces. Divide 23 Rolls-Royces into 2/3 + 1/6 + 1/8. The answer erroneously asserts this works for n  -1 (mod 24). David Singmaster. A Middle Eastern muddle. 41 oil wells to be divided into 1/2 + 1/3 + 1/7. But then I ask if there are values other than 2, 3, 7, 41 which produce such a puzzle problem. There are 12 such quadruples. I recall seeing this when I was a student but I haven't relocated it. Appeared in my puzzle columns as follows. "Well, well, well." Brain Twister. Weekend Telegraph (27 Feb 1988) xv (misprinted), (5 Mar 1988) (corrected) xv & (12 Mar 1988) xv. Reprinted, with no title, in: The Daily Telegraph Braintwisters No. 1; Pan Books, London, 1993; with Barry R. Clarke, Rex Gooch and Angela Newing. Prob. 25, pp. 27, 76 & 117. "A Middle Eastern muddle." The Puzzle Box. Games & Puzzles 12 (Mar 1995) 1819 & 13 (Apr 1995) 40. John P. Ashley. Arithmetickle. Arithmetic Curiosities, Challenges, Games and Groaners for all Ages. Keystone Agencies, Radnor, Ohio, 1997. P. 8: Omar divides 17 horses among 3 sons. The Answer says: "It was a great solution, but it was not correct mathematics. The sum of the fractional parts: 1/2, 1/3 and 1/9 do not add up to 1 but to 17/18. Therefore, each of the heirs got a bit more than the will intended." David Singmaster. The seventeen camels and the thirteen camels. Draft paper written in Dec 2000. This discusses the classic 17 camels problem and the 13 camels problem given by Always (1971) finding all solutions for two, three or four sons. 7.G.2.POSTHUMOUS TWINS, ETC.  tA man dies, leaving a pregnant wife and a will explaining how his estate is to be divided between the wife and a son or a daughter. The wife produces twins, one boy and one girl. How is the estate to be divided? The most common version has son : wife = wife : daughter = 2 : 1 given in the will and derives son : wife : daughter = 4 : 2 : 1. I will denote this as the "usual form". Other proportions cited by Smith are: 4 : 3 : 2; 2:2: 1; 9 : 6 : 4. Alcuin & BR proceed by dividing the estate in half first. See Brooks for some odd versions. See Tropfke 655.  tMoritz Cantor. Vorlesungen Gber Geschichte der Mathematik. Vol. 1, 3rd ed., Teubner, Leipzig, 1907. Pp. 561563 sketches the history of this problem. This is the basis of Smith's discussion below. This asserts that the problem is based on Roman Lex Falcidia of -40 which required that at least  of an estate should go to the legal heir. He says it first appears in the works of Celsus and quotes Julianus. He also cites Caecilius Africanus (c100) and Julius Paulus (3C). Describes the common case and says that if the will was invalidated, then only the children would inherit. D. E. Smith. Op. cit. in 3. Based on Cantor, cites Lex Falcidia, Celsus, Julianus, Africanus. He cites 22 medieval references, including Vander Schuere and Recorde of those below. See also: Sanford 218-219; H&S 87-88; A&N 24-26. Juventius Celsus. De istituzione uxoris et postumi et postumae. c75. ??NYS cited by Smith. Julianus cites him. Salvianus Julianus. c140. Lex 13 principio. Digestorum lib. 28, title 2. ??NYS quoted by Cantor and cited by Buteo & Smith. Cantor says this reports a case, though the quoted text isn't very specific. Usual form. Julianus cites Celsus. Caecilius Africanus. c150. Lex 47 1. Digestorum lib 28, title 5. ??NYS cited by Cantor & Smith. Cantor says this refers to a case. Julius Paulus. 3C. Lex 81 principio. Digestorum lib. 28, title 5. ??NYS cited by Cantor & Smith. Cantor says this refers to a case. Alcuin. 9C. Prob. 35: Propositio de obitu cujusdam patrisfamilias. Problem of posthumous twins. Ratios are 3 : 1 for son : mother and 5 : 7 for mother : daughter. He takes half the estate and shares it 3 : 1 and then the other half is shared 5 : 7. This gives 9:8 : 7. Ahrens, A&N, p. 26, suggests 15 : 5 : 7, which is the result of the usual Roman process. BR. c1305. No. 91, pp. 110-111. Son : wife = wife : daughter = 3 : 2. Divides in halves, as in Alcuin, and divides each half as 3 : 2, giving son:wife : daughter = 3 : 5 : 2. Gherardi?. Liber habaci. c1310. P. 145. Usual form. Gherardi. Libro di ragioni. 1328. P. 37. Son : wife = 3 : 1; wife:daughter = 2 : 1. Divides as 6 : 2 : 1. Lucca 1754. c1330.  x  thF. 60r, pp. 136-137. Posthumous triplets, 2 boys and a girl with usual ratios. He divides in proportion 4 : 4 : 2 : 1 for boy : boy : mother: girl. F. 83r, pp. 200-201. Posthumous twins. Usual form.  x  tPseudodell'Abbaco. c1440. Prob. 100, p. 85 with plate on p. 86. Posthumous twins. Usual form. I have a colour slide of this. AR. c1450. Prob. 209, pp. 97, 176, 223. Man has son, wife and two daughters and gives the usual ratios, hence divides in the proportion 4 : 2 : 1 : 1. Muscarello. 1478. Ff. 75r76r, pp. 189191. Posthumous twins. Usual form. Wagner. Op. cit. in 7.G.1. 1483. Pp. 73-75 & 202-203. Usual will, but wife produces a son and two daughters. Divides as in AR. Chuquet. 1484. Prob. 205. English & discussion in FHM 205. Usual form. FHM say it "goes back to the Roman emperor and legislator Justinian" and quotes Recorde. HB.XI.22. 1488. P. 44 (Rath 247). Posthumous twins. Pacioli. Summa. 1494.  x  thF. 158r, prob. 80. Usual posthumous twins. Then says that Nofrio Dini of Florence, a respectable merchant in Pisa, at the shop of Giuliano Salviati, told him about such a will on 16 Dec 1486. After a bequest to the church, there was an estate of 800 to be divided. If a son was born, the mother was to get 400; if a daughter, the mother was to get 300. Twins were born and he says to divide as 3 : 3 : 5. Says one can deal similarly with similar problems. Of the Biographical Sources listed in Section 1, Taylor, p. 149 & Fennell, p. 11, mention this problem. F. 158v, prob. 82. Selling a pregnant cow which bears twins. Gives some rules which determine the relative values.  x  tBlasius. 1513. F. F.ii.v: Quarta regula. Dying man with pregnant wife. If she has a son, he gets 3/5 and the mother and the church get 1/5 each. If she has a daughter, the daughter and the mother get 2/5 each and the church gets 1/5. She has a son and a daughter. He divides in proportion 3 : 2 : 2 : 1, but gives no reason. Offhand, I would think that 1/5 should go to the church since this is specified in either case and then the remaining 4/5 should be divided in the proportion 3:1 : 1, giving overall proportions of 12 : 4 : 4 : 5. He says one can similarly deal with two sons or two daughters or two sons and one daughter. Tagliente. Libro de Abaco. (1515). 1541. Prob. 100, ff. 50v51r. Usual form. Ghaligai. Practica D'Arithmetica. 1521. Prob. 23, f. 65v. Usual posthumous twins. Riese. Rechnung. 1522. 1544 ed. p. 80; 1574 ed. p. 54v. Father leaves a widow, a son and two daughters. Divides as in AR. Tonstall. De Arte Supputandi. 1522. Quest. 16, pp. 146147. Usual form. Then considers 3 sons and 2 daughters! Cardan. Practica Arithmetice. 1539. Chap. 66, section 87, ff. GG.vi.v GG.vii.r (p.164). Posthumous twins. Son:wife = 4:1; wife:daughter=2:1; divides as 8:2:1 Giovanni Sfortunati. Nuovo lume. Venice, 1545. F. 58v. ??NYS described by Franci, op. cit. in 3.A, p. 38. Franci's discussion is about several extended versions, but it seems to indicate that Sfortunati deals with a hermaphrodite. Recorde. Second Part. 1552. Smith, op. cit. in 3.A, p. 69, quotes 1558, fol. X 8 (??NYS). 1668, pp.289293: A question of a Testament. Man with fortune of 3600 and a pregnant wife makes a will and dies. If she has a son, the son get  and she gets @; if she has a daughter, she gets  and the daughter gets @. "If some cunning lawyers had this matter in scanning, they would determine this testament to be quite voyde, and so the man to die intestate, because the testament was made unsufficient." (The 1668 has identical wording, except it uses 'void' and 'insufficient'.) He divides in the proportion 9 : 6 : 4. Tartaglia. General Trattato. 1556. Book 12, art. 3541, pp. 199v200v.  x  thArt. 3540 give various ratios S : M = son : mother and M:D = mother : daughter, then divides in the usual way to get the proportions s : m : d such that s/m = S/M and m/d= M/D. E.g. when S : M = 2 : 1 = M : D, then s:m:d=4:2:1. Art. 35. S : M = 2 : 1; M : D = 2 : 1. Art. 36. S : M = 5 : 3; M : D = 1 : 1. Art. 37. S : M = 2 : 1; M : D = 1 : 1. Art. 38. S : M = 2 : 1; M : D = 2 : 1, in a different context than Art. 35. Art. 39. S : M = 2 : 1; M : D = 3 : 1. Art. 40. S : M = 3 : 1; M : D = 2 : 1. Art. 41. S : M = 2 : 1; M : D = 2 : 1, but quadruplets are produced two sons and two daughters. He divides in proportion 4 : 4 : 2 : 1 : 1.  x  tButeo. Logistica. 1559.  x  thProb. 60, pp. 264266. Usual form. Cites Julianus. Prob. 12, pp. 341342. Selling a pregnant cow, where the value depends on the sex of the calf. Cow + daughter is worth 40, while cow + son is worth 45. This is insufficient to determine the relative values, but he then adds excessive information: C = 3D = 2S. The cow produces twins one son and one daughter.  x  tGori. Libro di arimetricha. 1571.  x  thFf. 75r-75v (pp. 8384). Usual form. F. 75v (p. 84). Posthumous quintuplets divides in same proportions, though there is some confusion in the text of the solution.  x  tJacob Vander Schuere. Arithmetica, oft Reken-const. G. Kooman, Haarlem, 1600. ??NYS. [Smith, Rara, 421-423.] F. 98 is quoted in Smith, op. cit. in 3, p. 69, note 7. Posthumous triplets: boy, girl and hermaphrodite. Divides in proportion 12 : 4 : 2 : 7 = son : wife : daughter : hermaphrodite. Smith doesn't give the original ratios, but they were probably son : wife = 3 : 1, wife : daughter = 2 : 1. Schott. 1674.  x  th$nigma VII, pp. 559560. Usual form. $nigma X, p. 560. Son : wife = wife : daughter = 2 : 1, but he interprets this as the son getting 2/3 of the estate, the wife getting 2/3 of the rest with residue going to the daughter, leading to son : wife : daughter = 6 : 2 : 1.  x  tW. Leybourn. Pleasure with Profit. 1694. Prob. 15, pp. 3940. Posthumous triplets: boy, boy, girl. Usual ratios. Divides 4 : 4 : 2 : 1. Ozanam. 1725. Prob. 24, 1725: 179. Prob. 4, 1778: 187188; 1803: 185185; 1814:160-161; 1840: 83. 1725 gives just posthumous triplets two girls and a boy. He divides 4 : 2 : 1 : 1. Montucla does usual form, then remarks that one could have posthumous triplets, e.g. two sons and a daughter, and that he thinks that the will would be declared legally void. Les Amusemens. 1749. Prob. 54, p. 186. Usual case. Vyse. Tutor's Guide. 1771? Prob. 19, 1793: pp. 156157; 1799: p. 167 & Key p. 209. Usual case. Dodson. Math. Repository. 1775. P. 13, Quest. XXXIV. Usual case. Charles Hutton. A Complete Treatise on Practical Arithmetic and Bookkeeping, .... New edition, corrected and enlarged by Alexander Ingram. [c1780?] G. & J. Ross, Edinburgh, 1804. [BMC earliest entry is 7th ed., 1785, then 14th ed., 1815.] Prob. 53, p. 137. Usual form, but states that the mother thereby loses 2400  compared to the case of just having a girl. What would she have got if she had only had a son? Answer is 2100  which assumes the usual 4 : 2 : 1 division for the case of twins. Vinot. 1860. Art. XLI: Testament ! interpr)ter, pp. 6162. First gives usual solution. The says the problem is not serious because French legislation gives a solution. Since the wife receives at least a third in either case mentioned by the husband, she must receive a third in any case. The author then suggests the rest be divided equally among the children if more than one is born. Edward Brooks. The Normal Mental Arithmetic A Thorough and Complete Course by Analysis and Induction. Sower, Potts & Co., Philadelphia, (1858), revised, 1863. Further revised as: The New Normal Mental Arithmetic: A Thorough and Complete Course by Analysis and Induction; Sower, Potts & Co., Philadelphia, 1873. Several examples in an unusual context.  x  th1863 p. 128, no. 15; 1873 pp. 167168, no. 5. Man left $26,000 to wife, son and daughter. If the daughter dies before coming of age, the widow gets ; if the son dies before coming of age, the widow gets ; what happens if all live? Unusually for this book, this problem has a remark which says the division should be in the proportion son:wife:daughter=9:3: 1. 1863 p. 128, no. 18; 1873 p. 168, no. 9. Man with children abroad and wife at home. If the son does not return, the widow gets A; if the daughter does not return, the widow gets @; both return and it is found that the son gets $3000 more than the daughter. What was the estate? 1863 p. 128, no. 19; 1873 p. 168, no. 10. A, B, C are thinking of buying a farm. They agree that if A and B buy it, then A pays 2/5 and if B and C buy it, then B pays 2/5. All three buy it together and C is found to pay $500 more than A. What was the cost?  x  tSusan Cunnington. The Story of Arithmetic. Swan Sonnenschein,, London, 1904. Prob. 11, p. 212. Usual form. Asserts it is a Roman problem of +300, but gives no references. Collins. Fun with Figures. 1928. Then he put in his other foot, pp. 236237. Usual form. He adds: A further complication triplets, two boys and a girl. "The easiest way to find out is to let the lawyers decide it, and it is the one best bet that they will get it all." The Little Puzzle Book. Op. cit. in 5.D.5. 1955. Pp. 5354: The judge's dilemma. Ratios are 2 : 1 for son : mother and 3 : 1 for mother : daughter. Divides as 6 : 3 : 1. 7.H.DIVISION AND SHARING PROBLEMS CISTERN PROBLEMS  tSee Tropfke 578. The earliest sources in this group include what I call 'assembly problems'. In these, there are several processes which constitute a unit of work. The rates for the processes are given and one has to determine the number of units which can be done in a day (or how long some number of units will take). See the Babylonian examples below and: Chiu Chang Suan Ching; Heron; Metrodorus 134, 136; Bakhshali MS; BR 97, 98; AR 57, 75; Muscarello; Borghi; Riese; Cardan; Tartaglia; Pike; Treatise, 1850; Chambers; Bullen; Pearson. I am indebted to Eleanor Robson for the Old Babylonian examples. She has also provided most of the references to the source material which I not yet seen. The dating of these examples is generally pretty vague. Note. Cistern problems with two pipes have the same form as meeting problems, cf. 10.A. NOTATION: (a, b, c, ...) means that different pipes, etc. can do the job in a, b, c, .... How long for all together? Negative values indicate outlets. l  t""  HaHbH... H30H20HHHHLucca 1754; Wingate/Kersey; Wells; H20H16HHHHTartaglia H20H15HHHHBenedetto da Firenze H18H12HHHHCalandri, c1485 H16H10HHHHDe Morgan, 1831 H15H12HHHHCalandri, 1491; Tagliente; De Morgan, 1836; H14H12HHHHHuttonRutherford H13H10HHHHVyse; Bonnycastle; Colenso H12H7HHHHPike H10H8HHHHPseudodell'Abbaco H10H7HHHHMuscarello H10H6HHHHBR 99 H10H15HHHHD. Adams, 1835 H9H12HHHHLes Amusemens H8H6HHHHCalandri, c1485; Hutton; Mittenzwey H7H5HHHHDe Morgan, 1836; Sonnenschein & Nesbit H6H4HHHHBenedetto da Firenze; Calandri, 1491; Tagliente; Gori H5H4HHHHPseudodell'Abbaco H4H3HHHHLucca 1754; D. Adams, 1835; Burnaby H4H1HHHHHeron; BR 98 H4H11HHHHCalandri, 1491; Tonstall H14/4H5/2HHHHLacroix H3H2HHHHGherardi? H3H9HHHHBR 70 H2H3HHHHButeo H60H30H20HHHMeichsner H55H45H30HHHPike H50H40H25HHHHutton, c1780?; Eadon; Colenso H27H15H12HHHD. Adams, 1801 H24H12H8HHHButeo H20H15H12HHHUnger 520 H20H15H10HHHCalandri, c1485 H18H12H6HHHCalandri, c1485 H16H12H10HHHSimpson H16H12H8HHHCalandri, c1485 H15H12H10HHHBenedetto da Firenze; Calandri, 1491 (twice); HHHHHHDictator Roffensis; H12H10H9HHHSonnenschein & Nesbit H12H10H8HHHMittenzwey H12H10H6HHHCalandri, c1485 H12H9H6HHHBorghi; Ozanam H12H8H4HHHMetrodorus 131 H12H8H10HHHUnger 521 H10H9H8HHHMilne H10H8H4HHHPacioli H10H5H4HHHPacioli; Tartaglia H9H7H2HHHPike H9H6H4HHHMittenzwey H8H6H4HHHCalandri, 1491; Tonstall H8H6H3HHHPacioli H7H5H6HHHKing H7H5H4HHHAR 70 H7H5H3HHHGherardi H6H5H4HHHFibonacci H6H4H3HHHChuquet H6H4H2HHHMetrodorus 135; OzanamMontucla H6H4H4HHHSonnenschein & Nesbit H6H3H1HHHFaulhaber H5H4H3HHHLucca 1754; Gori; Les Amusemens H5H3H2HHHCalandri, c1485 H4H3H2HHHGherardi?; AR 98; Wagner; Faulhaber H4H2H1HHHGori H3H8/3H12/5HHHNewton; Dodson; Eadon; Colenso H3H2H1HHHMetrodorus 133; Anania(?); alKarkhi; BR 64; AR51, 97;  tCalandri, 1491; Blasius; Tonstall; Riese; Vyse; King l  t"" H3H1H2/5HHHMetrodorus 132 H5/3H1/2H1/3HHHChaturveda H1H3/4H1/2HHHWingate/Kersey H1H1/2H1/4HHHAR 281; Tonstall H1/2H1/3H1/4HHHColumbia Alg.; Pike H1/2H10/7H7/3HHHWingate/Kersey H80H40H20H10HHD. Adams, 1801 H72H60H20H12HHLevi ben Gershon H27H24H9H6HHFibonacci H6H8H9H12HHRecorde H6H5H4H3HHBartoli H6H5H3H2HHMuscarello H6H4H3H2HHW. Leybourn H4H3H2H1HHMetrodorus 130; Fibonacci; Tonstall H4H3H2H1/2HHMetrodorus 7; BR 65; van Etten; Wingate/Kersey; H4H3H2H1/4HHSchott; Ozanam H1H1/2H1/3H1/6HHBhaskara II H1H1/2H1/4H1/5HHChaturveda H1/2H1/3H1/4H1/5HHMahavira H1/2H1/4H1/5H1/6HHSridhara H H5H3H5/2H1H1/3HChiu Chang Suan Ching H4H3H2H4H6HBR 116 H3H2H3H4H5Hdella Francesca H1/2H1/3H1/4H1/5H1/6HColumbia Alg. H1/2H1/3H1/5H1/7H1/9HBR 25 H H6H5H4H3H2H1Bartoli H4H3H2H3H4H5 della Francesca H3H2H1H3/4H4H5 Cardan H3H2H1H2H3H4 Pacioli H12H10H8H6H3H45 6 Bullen; Treatise, 1850  t General solution see: Levi ben Gershon; Wells; Newton; Simpson; Dodson; Bonnycastle; Hutton; Lacroix; DeMorgan; Bourdon; Young; Mittenzwey; Milne. The earliest forms derive joint rates from individual rates. Deriving individual rates from joint rates seems to begin in the 14C. NOTATION: (A, x) in B means the first can do it in A and the first and second together can do it in B. How long would it take the second? For such problems, see: BR; Gherardi; Pseudodell'Abbaco; AR; Treviso Arith.; Chuquet; Calandri, 1491; Tonstall; GemmaFrisius; Tartaglia; Buteo; Wingate/Kersey; Wells; Simpson; Euler; Vyse; Dodson; Ozanam-Montucla; Bonnycastle; Pike; Bullen; Eadon; Hutton,1798?; Bonnycastle,1815; Jackson; NutstoCrack; D. Adams, 1835; Family Friend; Treatise,1850; Colenso; Docharty; Thomson; Brooks.  t (50, x)in 36Gherardi (48, x)in 24Docharty (36, x)in 30Gherardi (36, x)in 24Docharty (30, x)in 12Dodson; Bonnycastle; Hutton, 1798?; Nuts to Crack (20, x)in 60Silvester (20, x)in 14Gemma Frisius (20, x)in 12Wingate/Kersey; Wells; Euler; Dodson; Pike; Bonnycastle,1815; Mittenzwey (20, x)in 8Treviso Arith. (18, x)in 11Vyse (35/2, x)in 40Docharty (gives a negative x!) (16, x)in 10Treatise, 1850 (15, x)in 18Thomson (gives a negative x!) (15, x)in 10Treatise, 1850 (13, x)in 9AR 76 (13, x)in 8Pike (12, x)in 3Family Friend, 1849 (10, x)in 7Colenso ( 9, x)in 5Pseudodell'Abbaco ( 8, x)in 5Buteo; Eadon ( 7, x)in 5D. Adams, 1835 ( 5, x)in 15/8BR 67 ( 3, x)in 4/3BR 66 ( 3, x)in 9/2BR 69 (negative value!) (9, x)in 9/2BR 68 (80, 60, x)in 30Tartaglia (44, 32, x)in 16Eadon (40, 30, x)in 15Calandri, 1491; Tonstall; Wingate/Kersey (37, 23, x)in 15Pike (34, 24, x)in 12Vyse (17/2, 21/4, x)in 6/5Treatise, 1850 (8, 6, x)in 3Brooks (5/2, 9/4, x)in 1Treatise, 1850  t For the general solution of: (x, y) in A, (y, z) in B, (x, z) in C, see: della Francesca; Simpson; Euler; OzanamMontucla; Bonnycastle; Hutton; De Morgan, 1836; Colenso; Singmaster. For examples of this form, see also: Muscarello; Dodson; D. Adams, 1835; Docharty; Todhunter; Sonnenschein& Nesbit. This is a form of the type III problem in Section 7.R.1, where the inverses of the variables are used. Singmaster asks how to choose A, B, C so that x, y, z and the time for all three together are all integers the case with data 20, 15, 12 is by far the simplest example and none of the other examples have this property. lt" HAHBHC H60H4H40Colenso H30H20H15AR 182 H20H15H12Docharty; Todhunter; Singmaster H15H12H10della Francesca H14H12H21/2Colenso H10H9H8Simpson; Euler; Dodson; OzanamMontucla; Bonnycastle; HHH Hutton; Docharty; Vinot; H9H8H6Sonnenschein & Nesbit H5H4H3Muscarello H4H6H5D. Adams, 1835  t Vyse, Docharty and Thomson are the only examples I have seen with four people and you know how long it takes each set of three. Fish has five workers and you know how long each four take. If you use the reciprocals of the times, then these are like type III problems in 7.R.1. That is, if A, B, C, D take A, B, C, D days, their rate of work is a = 1/A per day, etc. Then saying that A, B, C can do it in d4 days becomes a + b + c = 1/d4, etc. For problems where the combinations involve one tap or worker working only part of the time that the other does, see: Fibonacci; Gherardi; Chuquet; Cardan; Buteo; Pike; Jackson; Treatise, 1850; Colenso; Young; Chambers; Brooks; Andr); Sonnenschein & Nesbit. For problems like (x,x/2,x/3) in 2, see: di Bartolo; Buteo; Todhunter. For problems like (x, x-5) in 12, which lead to quadratic equations, see: Di Bartolo; Buteo; Tate; Todhunter; Briggs & Bryan. Sonnenschein & Nesbit has a version where pumps can work at half or full power. I have included a few direct rate problems as comparisons these usually involve money see: Bakhshali; Chaturveda; Pike; Chambers. See Clairaut for the use of this context to discuss negative solutions. See Smith, op. cit. in 3. See also 7.E & H&S 69-71. 5.W.1 can be viewed as parodies of this problem. COMPARISON of assembly and cistern problems. Consider the cisterntype problem (a1, a2 , ...). In the unit of time, the pipes do 1/a1, 1/a2, ... of the work, so all together they do S = $ 1/ai per unit time and so the whole job takes time 1/S. In an assemblytype problem, we can do ai units of process i per unit of time. Hence it takes 1/ai time to do one unit of process i. If each process has to be done the same number of times, then it takes S = $ 1/ai time to do a unit of work and so 1/S units can be done in a unit of time. In the Babylonian problems, the unit of work may require varying amounts of the different units. If the unit of work requires bi units of process i, then we take S = $ bi/ai. Hence the problems are mathematically the same, though the formulations are different.  tYBC 7164. Old Babylonian problem tablet at Yale, problems 6 & 7, c-1700? Transcribed, translated and commented on in Neugebauer & Sachs, op. cit. in 7.E, 1945, pp. 8188 & plate 10 & photo plate 35. On pp. 148149, a linguistic analysis says it probably comes from Larsa, in southern Mesopotamia.  x  thProblem 7. A canal has to be cleaned to 3 kI deep. A man can clear 20 g1n of silt from the top kI in a day or he can clear 10 g1n from the lower level in a day. How much can he clear in a day? Here a1 = 20, a2 = 10, and we can take b1 = 1, b2Ġ= 2, because the lower level is twice as thick as the upper level. Problem 6. This is the same, but with depth 4 kI divided into three levels with the rate of doing the bottom level from 3 to 4 deep being only 7 g1n per day. So we just add a3 = 7 and b3 = 1 to the previous problem.  x  tBM 85196. Late Old Babylonian tablet in the British Museum, prob. 16, c-1700?. Transcribed, translated and commented on by O. Neugebauer; Mathematische Keilschrifttexte II; Springer, Berlin, 1935, pp. 45+, 49, 56+ ??NX. [See 6.BF.2 for another problem from this tablet.] But Neugebauer was not able to make sense of it until he saw the above problems, so it is reconsidered in Neugebauer & Sachs, pp.88-90. Robson says it is definitely from Sippar (middle Mesopotamia) and cites ThureauDangin; Revue d'Assyriologie 32 (1935) 1+ for another publication of the text, ??NYS. This problem and those of YBC 7164 are more recently discussed by Marvin A. Powell; Evidence for agriculture and waterworks in Babylonian mathematical texts; Bulletin on Sumerian Agriculture 4 (1988) 161172, ??NYS Like problem 6 above, with each level of depth 1 kI and rates of 20, 10, 6A g1n per day. In Spring 1994, I mentioned the assembly problems from the Chiu Chang Suan Ching (see below) in a lecture at Oxford. Eleanor Robson told me that such problems occur in Old Babylonian times and she sent me details, including the above references, and later provided more details and references. She described four further examples, without specific dates, and the next four examples are simplified from her letter. The simplifications are basically to avoid use of coefficients giving the number of bricks per unit of weight, etc. Haddad 104, c1770. Tablet from Tell Haddad, near Baghdad, found in the destruction layer from when Hammurabi conquered the site usually dated at 1762. The tablet is in Baghdad. See: Farouk alRawi & Michael Roaf; Ten Old Babylonian mathematical problems from Tell Haddad; Sumer 43 (1984) 175218.  x  thProb. ix Making bricks. One man can dig 1/3 sar of earth in a day, or he can mix 1/6 sar or he can mould 1/3 sar into bricks. If 1 sar makes 1620 bricks, how many bricks can a team of three make in a day? For one man, we get S=3+6+3, so he can process 1/12 sar per day, or 135 bricks, so three men can make 405 bricks. Prob. x Carrying earth to make bricks. Same problem as the previous, but the earth must be carried 5 nindan from the digging site to the works. The amount one man can carry in a day is given somewhat cryptically. The simplest interpretation is that one man can carry 1/3 sar of earth over the 5 nindan in a day, but there still are three workers in the group. Here we get S = 3 + 3 + 6 + 3, so one man can process 1/15 sar per day or 108 bricks and three men make 324 bricks.  x  tYBC 4669. This and the following tablet are in the same hand, but have no provenance. See Neugebauer, vol. III, pp. 2829 & plate 3, ??NYS. Reverse, col. 3, lines 717, c1800 Demolishing walls. A man can knock down 1/15 sar of wall in 1/5 of a day and he can carry away 1/12 sar in a day. How much wall can he demolish and carry away in a day; and what part of the day is devoted to each task? Here a1 = (1/15)/(1/5) = 1/3, so S = 3 + 12 and he can do 1/15 sar per day. YBC 4673. See Neugebauer, vol. III, pp. 30 & 32 & plate 3, ??NYS. Obverse, col. 2, lines1018, c1800 Constructing a pile of bricks. A man can carry 1/18 sar of earth (bricks??) in a day. He can pile up 1 sar of bricks in 14 2/5 days. If a sar makes 5184 bricks of this size, how many bricks can he carry and pile up in a day? Chiu Chang Suan Ching (Jiu Zhang Suan Shu). c-150? Chap. VI.  x  thProb. 22, p. 67. Man can do two processes at rates of 38 in 3 days and 76 in 2 days. How many of both together can he do in one day? Answer is given as 25 but Vogel's note on the calculation shows 25@ was meant, and this is erroneous the correct answer is 9. The error arises from taking 38 and 76 as rates per day. Prob. 23, p. 67. Three processes at rates 50, 30, 15 per day, how many together in a day? Correct answer, 8@, is obtained. (Arrow shafts, arrow feathering, arrow heading.) Prob. 25, p. 68. Three processes at rates 7, 3, 5 per day, how many together in a day? Correct answer, 105/71, is obtained. Prob. 26, pp. 68-69. Cistern: (1/3, 1, 5/2, 3, 5). Correct answer. Vogel says this is the first appearance of the problem.  x  tHeron (attrib.). c150.  #  )# (Peri Metron). In: J. L. Heiberg, ed.; Heronis Alexandrini Opera Quae Supersunt Omnia, vol. V; Teubner, Leipzig, 1914; reprinted 1976, pp. 176-177. Greek and German texts.  x  thProblem 20:  )#%% /%) #% (Metresis xisternas) [Vermessung einer Zisterne]. (1,4) in hours, but he computes 1 + 4 = 5 and then sets the cistern = 12 ft and computes 12/5 as the number of hours. See BR, c1305, prob. 98, for the explanation of this. Problem 21: 3%   )# %% (Allos e metresis) [Die Vermessung in anderer Weise]. + 1/7 - 1/11, how long to make 100? He treats the - as + and gets the correct answer for that case, though Heiberg says the calculation is senseless.  x  tSmith, History II 538, quotes from Bachet's Diophantos, implying a date of c275, citing the 1570 edition with Fermat's notes, but Smith's citation is to the part of Bachet taken from Metrodorus! It is Art. 130 of Metrodorus. Sanford 216 also cites Diophantos, but her discussion is based on Smith's AMM article (op. cit. in 3), which is the basis of the section in Smith's History containing Smith's quote. The problem is nowhere in Heath's edition of Diophantos. However, Tropfke 578 gives a reference to the Tannery edition of Diophantos, vol. 2, p. 46 ??NYS. Metrodorus. c510. 8 cistern-type problems.  x  thArt. 7, pp. 30-31. "I am a brazen lion." (2, 3, 4, 1/2), where 6 hours is counted as 1/2 day, i.e. a day has 12 hours. Art. 130, pp. 96-97. (1, 2, 3, 4). Art. 131, pp. 96-97. (4, 8, 12). Art. 132, pp. 96-97. "This is Polyphemus, the brazen cyclops." (3, 1, 2/5). Art. 133, pp. 96-99. (1, 3, 2). Art. 134, pp. 98-99. Three spinners can do 1, 4/3, 1/2 unit per day, how long for all three to do one unit? Art. 135, pp. 98-99. "We three Loves" (or Cupids). (2, 4, 6). Art. 136, pp. 98-101. 'Brickmakers.' Three brickmakers can make 300, 200, 250 per day. How long for all three to make 300?  x  tBakhshali MS. c7C. Kaye I 49-52 discusses several types, e.g. first gives 5/2 dinars in 3/2 days; next gives 7/2 in 4/3; third gives 9/2 in 5/4; how long for all three to give 500 dinars? (= Kaye III 192, ff. 21v22r). Kaye III 191 has three rates of 1/(1/3),1/(1/2),3/5 how long to give 100? I 51 (= III 233234, ff. 44v44r) is an example with an income, some capital and three rates of expenditure. On I 50 (= III 234235, ff.44r43v) is an example with an income, some capital and seven rates of expenditure! Anania Schirakatzi (= Ananias of Shirak). Arithmetical problems. c640. Translated by: P.Sahak Kokian as: Des Anania von Schirak arithmetische Aufgaben; Zeitschrift fGr d. deutsch?sterr. Gymnasien 69 (1919) 112117. See 7.E for description. Prob. 24 is a cistern with pipes: (1, 2, 3), but he gives the answer: 1/4 + 1/6 + 1/16 + 1/18, which = 77/144, which is close to the correct answer of 6/11. No working is shown and I am unable to see how 77/144 can arise, even allowing for a possible misprint. H&S 70 says the cistern problem appears in Alcuin, 9C, but the only possible problem is a trivial problem (8: Propositio de cupa) which mentions a barrel. Mahavira. 850. Chap. VIII, v. 32, p. 266. Cistern: (1/2, 1/3, 1/4, 1/5). Chaturveda. 860. Commentary on Brahma-sphuta-siddhanta, chap. XII, sect. 1, art. 9. InColebrooke, p. 282.  x  thCistern: (1, 1/2, 1/4, 1/5). (Datta & Singh, I, 234 and others cite this as Brahmagupta.) Bestowing alms: (1/3, 1/2, 5/3).  x  tSridhara. c900. V. 69, ex. 91, pp. 55-56 & 95. Cistern: (1/2, 1/4, 1/5, 1/6). al-Karkhi. c1010. Sect II, no. 15-16, p. 83. Cistern: (1, 2, 3). No. 16 asks how often the cistern will be filled in 5 days. Bhaskara II. Lilavati. 1150. Chap. IV, sect. II, v. 95. In Colebrooke, p. 42. Cistern:(1,1/2,1/3, 1/6). (Datta & Singh, I, 234, erroneously say this is the same problem as Brahmagupta, i.e. Chaturveda.) Fibonacci. 1202.  x  thP. 182 (S: 279280): De Leone et leopardo et urso [On the lion and leopard and the bear]. Lion, leopard and bear eating a sheep: (4, 5, 6). P. 182 (S: 280): De duabus navibus ... [On two ships ...] is in 10.A. P. 183 (S: 281): cistern problems (1, 2, 3, 4) & (6, 9, 24, 27). Pp. 183-186 (S: 282285) several problems with water butts having different size openings at different heights. E.g., pp. 183-184 has four openings at 1/4,2/4,3/4, 4/4 of the way down, which could drain the whole butt in 4,8,12,16 days. How long to drain a full butt with all holes open? Answer:7267/2275 days.  x  tBR. c1305.  x  thNo. 25, pp. 44-45. Ship with 5 sails: (1/2, 1/3, 1/5, 1/7, 1/9). Vogel says this is the first example of the formulation of a ship with sails. No. 64, pp. 88-89. Cistern, (1, 2, 3). No. 65, pp. 88-91. Cistern 'I am a noble lion', (2, 3, 4, 1/2). =Metrodorus 7. No. 66, pp. 90-91. Cistern, (3, x) in 4/3. No. 67, pp. 90-93. Cistern, (5, x) in 15/8. No. 68, pp. 92-93. Cistern, (x, -9) in 9/2. No. 69, pp. 92-93. Cistern, (3, -x) in 9/2. No. 70, pp. 94-95. Cistern, (3, -9). No. 96, pp. 114-117. 3 cisterns of volumes 30, 60, 120 with pipes that fill them in 6,4, 3. Using all three pipes, how long to fill all three cisterns? No. 97, pp. 116-117. Cistern, + 1/7 - 1/11 to yield 100. This is Heron's prob. 21 and is done in the same way as though it were + 1/7 + 1/11, but one MS is worded so this is the correct method, as noted by Vogel. No. 98, pp. 118-119. Cistern, (1, 4). This is Heron's prob. 20, again noted by Vogel. The text says to set the cistern equal to 12 and then divides by 5 = 1 + 4. Vogel notes that this does not give the time, but 12/5 is the volume delivered by the smaller pipe. No. 99, pp. 118-119. Cistern, (6, 10). No. 116, pp. 132-133. Cistern, (2, 3, 4, -6, -4).  x  tGherardi?. Liber habaci. c1310 Pp. 143-144: Compangnia et viaggio. Baratti xvii. Three workers, (2, 3, 4). Ship with two sails, (2, 3). Levi ben Gershon. Maaseh Hoshev (= Maaseh hosheb) (Work of the Computer), also known as Sefer ha mispar (Book of Number). 1321 or 1322. ??NYS translation of the following by Shai Simonson, sent by David E. Kullman. "Question: A certain container has various holes in it, and one of the holes lets all the contents drain out in a given times. And so on for each of the holes. How much time will it take to empty the container when all the holes are opened? First, calculate how much drains from each hole in an hour, add them all up, and note the ratio to the full container. This ratio is equal to the ratio of one hour to the time it will take t empty the container when all holes are open." He then does (72, 60, 20, 12). Gherardi. Libro di ragioni. 1328.  x  thPp. 44-45: Volare una bocte. (3,5), but the 5 is half-way down the barrel while the 3 is at the bottom. P. 45: Ship with three sails, (3, 5, 7). Pp. 56-57: Uno chavaleri che vuole far fare uno pallagio. Three workers, (50, B) in 36, (50, B, C) in 30.  x  tLucca 1754. c1330. F. 59r, p. 134.  x  thShip with 3 or 2 sails: (3, 4, 5), (3, 4). Two couriers meeting: (20, 30).  x  tColumbia Algorism. c1350.  x  thProb. 66, pp. 87-88. Cask can be emptied in 1/2, 1/3, 1/4, 1/5, 1/6 of a day. (See also Cowley 399.) Prob. 141, p. 150. Same with 1/2, 1/3, 1/4.  x  tGiovanni di Bartolo. Certi Chasi. c1400. Copied by Maestro Benedetto (da Firenze), in Cod.L.IV.21, Biblioteca degli Intronati di Siena, 1463. Edited by M. Pancanti, Quaderni del Centro Studi della Matematica Medioevale, No. 3, Univ. di Siena, 1982. Cf Van Egmond's Catalog 189190 which doesn't mention this material.  x  thProb. 1, pp. 4-5. (x, x/2) in 5. Prob. 2-8, pp. 5-17 are more complex examples, often leading to quadratic equations, e.g. (x, x-5) in 12.  x  tBartoli. Memoriale. c1420.  x  thProb. 14, f. 76v (= Sesiano, pp. 140142 & 149, with facsimile of the relevant part of f.76v on p. 141. Cask with four taps: (3, 4, 5, 6). Prob. 33, f. 79r (= Sesiano, pp. 146 & 150). Six workers building a wall: (1,2,3,4,5,6). He correctly finds the total rate is 137 [/ 60], but then uses 36 instead of 60 "Now make 6 times 6, which is 36, because there are 6 workers." Sesiano describes this as fantasy.  x  tPseudodell'Abbaco. c1440.  x  thProb. 23, p. 32. (4, 5). Prob. 50, p. 49 with plate on p. 50. Ship with two sails, (8, 10). I have a colour slide of this. Prob. 62, p. 59. (9, x) in 5.  x  tAR. c1450. Prob. 51, 57, 70, 75, 76, 97, 98, 182, 281. Pp. 42, 44-45, 48-50, 58-59, 85-86, 128-129, 157, 160-161, 165-166, 175, 211-213, 221.  x  th51: cistern with three drains, (1, 2, 3), erroneously done see 97. 57: three mills, but gives amounts each can do per day. 70: three builders, (7, 5, 4). 75: three tailors, but gives amounts each can do per day. 76: (13, B) in 9. 97: cistern with three drains, (1, 2, 3), = Metrodorus 133. 98: ship with three sails, (2, 3, 4). 182: three scribes; (A, B) in 20, (A, C) in 30, (B, C) in 15, how long for each? 281: barrel with three taps, (1, , ).  x  tBenedetto da Firenze. c1465.  x  thPp. 90-91: ship with three sails (10, 12, 15). P. 91: cistern (4, 6). P. 91: two workers (20, 5).  x  t"The Treviso Arithmetic" = Larte de labbacho (there is no actual title). Treviso, 1478. Translated by David Eugene Smith, with historical commentary by Frank J. Swetz, as: Capitalism and Arithmetic; Open Court, La Salle, Illinois, (1987), improved ed., 1989. This is discussed in: D. E. Smith; The first printed arithmetic (Treviso, 1478); Isis 6 (1924) 310-331. Facsimile edition, from the copy at the Diocese of Treviso, with commentary booklet by Giuliano Romano, (Editore Zoppelli, sponsored by Cassa di Risparmio della Marca Trevigian, Treviso, 1969); updated ed., Libreria Canova, CalMaggiore 31, Treviso (tel:0422546253), 1995 [Swetz, p. 324, cites the 1969 ed.] See: www.calion.com/cultu/abbacho/abbacen.htm for a description of the book and how to order it. The facsimile has taken its title from the end of the opening sentence. Romano's commentary calls it: L'Arte dell'Abbacho. The text nowhere gives a publisher's name. Smith, Rara, pp. 37, says it was probably published by Manzolo or Manzolino, while Swetz, p. 26, specifies Michael Manzolo or Manazolus. Romano says it was published by Gerardus de Lisa. There was a copy in the Honeyman Collection, with title Arte dell'Abbaco and publisher [Gerardus de Lisa], who is described as the prototypographer at Treviso from 1471. The entry says only ten copies are known the web page says nine. F. 57r (pp. 162-163 in Swetz). Carpenters, (20, x) in 8. Muscarello. 1478.  x  thF. 58r, p. 161162. (B, C) in 3, (A, C) in 4, (A, B) in 5. There are two copying errors in the MS answers. F. 62v, pp. 169170. Four workers, (2, 3, 5, 6). Ff. 77r77v, pp. 192193. Three mills can do 9, 8, 5 per day. How long will it take them to do 6 and how much does each do? F. 77v, p. 193. Ship with two sails, (7, 10). F. 81r, p. 196. Cask with three spouts which can let out 6, 7, 8 per hour. How long will it take to empty a cask of 23?  x  tdella Francesca. Trattato. c1480.  x  thFf. 15r15v (6162). Basin with three inlets and three outlets, (2, 3, 4, 3, 4, 5). Plug the +4 pipe, which gives (2, 3, 3, 4, 5). English in Jayawardene. F. 127r (269). Three workers: A & B in 15; A & C in 12; B & C in 10. English in Jayawardene.  x  tWagner. Op. cit. in 7.G.1. 1483. Regel von einem Fass, pp. 114 & 224. Cask with three taps (2, 3, 4). Chuquet. 1484.  x  thProb. 21. English in FHM 204. Cistern emptying, (3, 4, 6). Prob. 53. English in FHM 209210. The first says: "If you help me 8 days, I will build it in 20". The second responds: "If you help me 10 days, I can do it in 15". How long for each alone? Prob. 54. Same as prob. 53 with parameters 5, 17; 6, 24.  x  tBorghi. Arithmetica. 1484.  x  thF. 106v (1509: ff. 91r91v). Three mills can grind 6, 9, 11 per day. How long to do 100? F. 109r (1509: ff. 91v92r). Ship with three sails, (6, 9, 12). (H&S 70 gives Latin and English.)  x  tCalandri. Aritmetica. c1485.  x  thF. 91r, p. 182. Ship with 2 sails: (12, 18). F. 91v, p. 183. Three men in prison: (6, 12, 18). (Tropfke 520 reproduces this in B&W.) F. 93r, p. 186. Emptying a cask: (6, 8). F. 95v, p. 191. Ship with three sails: (6, 10, 12). (Coloured plate opp. p. 120 of the text volume.) F. 96v, p. 193. Emptying a cask: (8, 12, 16). F. 97v, p. 195. Lion, wolf & fox eating a goat: (2, 3, 5). (Tropfke 581 reproduces this in B&W.) Ff. 98v99r, pp. 197-198. Furnace with 3 fires: (10, 15, 20).  x  tJohann Widman. Op. cit. in 7.G.1. 1489. (On pp. 131132, Glaisher mentions the following.) Ff. 136r-138v: Eyn fasz mit dreyen Czapfen; Von der Mulen; Leb, wolff, hunt; Schiff. (Cistern problem; 3 mills; lion, wolf, dog eating a sheep; ship with 3 sails.) Calandri. Arimethrica. 1491.  x  thF. 68v. Ship with two sails. (12, 15). Woodcut of ship with indeterminate number of sails. F. 69r. Cask with two taps. (4, 6). Woodcut of cask with two taps. F. 70r. Ship with three sails. (10, 12, 15). Same woodcut as on f. 68v. F. 70r. Cask with three taps. (4, 6, 8). Same woodcut as on f. 69r. F. 70v. Three masters build a house. (10, 12, 15). Woodcut of two builders. (H&S 70 gives Italian and English and says it also occurs in the Treviso Arithmetic (1478) [but that has a very different type!], Pacioli, Cataneo, Tartaglia, Buteo (1559), Clavius, Tonstall.) F. 70v. Three masters doing a job. (30, 40, x) in 15. F. 71v. Cistern. (4, -11). Woodcut of cistern. (Rara, 48 is a reproduction.) F. 72v. Lion, leopard & wolf eating a sheep. (1, 2, 3) days. Nice woodcut. (H&S 70 gives Italian and English, says there is a remarkable picture and says it occurs in Fibonacci [again, there it occurs in a different form] and Cataneo.)  x  tPacioli. Summa. 1494. See also Buteo.  x  thF. 99r, prob. 6. Building a house, (8, 10, 4). Says one can have more builders and it is similar to dog, wolf & lion eating a sheep. F. 99v, prob. 16. Three mills, (6, 8, 3) days. F. 99v, prob. 17(not printed). Three mills, (10, 5, 4) days.  x  t PART II.  x  thF. 66v. prob. 91. Cask with four taps. Volume above highest tap is 1/3 of the cask. Volume between highest and second highest is 1/4; volume between second and third highest is 1/5; volume between third highest and lowest tap is the rest of the cask. Each tap can empty the section just above it in 1, 2, 3, 4 days. How long to empty with all taps? He assumes the cask holds 60 so the rates are 20, 15/2, 12/3, 13/4 per day. Answer is 80/139 + 60/59 + 48/29 + 4, but he gives the sum as 7 245235/2959139. Clearly the denominator denotes 2959139=237829, but the correct sum is 7 58901/237829 and I cannot see how his expression relates to the answer. The answer is not 7+24/29+52/59+35/139, nor any similar expression that I can think of. Ff. 66v67r, prob. 92. Basin has inlets which fill it in 1, 2, 3 hours and outlet which empty it in 2, 3, 4 hours, i.e. (1, 2, 3, 2, 3, 4). How long to fill? He follows with remarks that all such problems can be done similarly. Cf della Francesca.  x  tBlasius. 1513. F. F.iii.r: Decimatertia regula. Three rivers can water a field in (1, 2, 3) days. Gets 13 1/11 hours for all three so he is using 24 hour days. Tagliente. Libro de Abaco. (1515). 1541.  x  thProb. 117, f. 58r. Ship with two sails (12, 15). Prob. 119, f. 58v. Cask with two taps (4, 6).  x  tTonstall. De Arte Supputandi. 1522.  x  thQuest. 26, pp. 157159. Three mills can do at rates of 18, 13, 8 per day. How long to do 24? Quest. 27, pp. 159161. Cistern, (1, 2, 3) and (4, 6, 8). Quest. 28, pp. 161162. Cistern, (1/4, 1/2, 1). Quest. 29, pp. 162163. Cistern, (4, 11). Quest. 32, pp. 164165. Four architects building a house, (1, 2, 3, 4) years. Says it is similar to a cistern problem. Quest. 33, p. 166. Three architects building a house, (30, 40, x) in 15.  x  tRiese. Die Coss. 1524.  x  thNo. 117, p. 55. Cask with three taps, (1, 2, 3). No. 118, p. 56. Three windmills can grind 20, 17, 15 per day. How long to do 24?  x  tCardan. Practica Arithmetice. 1539.  x  thChap. 47, ff. L.iii.r L.iii.v (pp. 7071). Simple example 5 mills grind 7, 5, 3, 2, 1 per hour, how long will they take to grind 500? Chap. 66, section 125, ff. kk.vi.r kk.vi.v (pp. 180181). Cask with four taps located at levels 1/3,1/3+1/4,1/3 + 1/4 + 1/6, 1 from the top and which empty their respective portions in 4, 3, 2, 1 hours. How long to empty the cask with all taps? Chap. 66, section 126 (misprinted 123), ff. kk.vi.v kk.vii.r (p. 181). Cistern:(1,2,3,-4, 5, 3/4).  x  tGemma Frisius. Arithmetica. 1540. (20, x) in 14 man & wife drinking a cask of wine. ??NYS Latin given in H&S, p. 71. Recorde. Second Part. 1552. 1668, pp. 329330: A question of water, the eighth example. (6, 8, 9, 12). Tartaglia. General Trattato, 1556, art. 74, p. 248v; art. 176-177, p.261v; art. 187-188, pp.262r-262v.  x  thArt. 74: 120 per 40 and 15  per 6 to do 120. Art. 176: (16, 20). Art. 177: (60, 80, x) in 30. Art. 187: 1 per 8, 1 per 6 and 1 per 3 to do 25. Art. 188: (10, 5, 4).  x  tButeo. Logistica. 1559.  x  thProb. 6, pp. 205206. Three mighty drinkers drinking an amphora of wine in (24, 12, 8) hours. Cites Pacioli. (H&S 71) Prob. 7, pp. 206208. Three architects build a house: (x,x/2,x/3) in 2 months. Says Pacioli gives (x, x+6, x+8) in 2 and solves it wrongly. Prob. 8, pp. 208209. Ship with two sails, (8, x) in 5. Prob. 61, pp. 266268. Cask with three taps 1/4, 2/3, 1 of the way down which could empty the whole cask in (6, 3, 3) hours. Prob. 62, pp. 268269. Cistern, (+2, 3).  x  tGori. Libro di arimetricha. 1571.  x  thF. 74v (pp. 82-83). Cistern empties in (4, 6) hours. Ship with three sails, (3, 4, 5) days. F. 77v (p.83). Lion, bear and wolf eating a sheep, (4, 2, 1) hours.  x  tJohann Faulhaber. Arithmetischer Wegweiser .... Ulm, 1614. ??NYS. A 1708 ed. is quoted in Hugo Grosse; Historische RechenbGcher des 16. und 17. Jahrhunderts; (1901); reprinted by Sndig, Wiesbaden, 1965, p. 120.  x  thNo. 91, p. 228: wolf, sheepdog and dog eating a sheep, (1, 3, 6). No. 92, p. 229: ship with three sails, (2, 3, 4).  x  tvan Etten. 1624. Prob. 83 (76): Du Lyon de Bronze pos) sur une fontaine avec cette epigraphe, pp. 94-95 (140). (2, 3, 4, 1/2) = Metrodorus, art. 7. Georg Meichsner. Arithmetica Historica. Hieronymus K?rnlein, Rotenburg/Tauber, 1625. No. 68, p. 209. ??NYS. Quoted in Hugo Grosse, op. cit. under Faulhaber, above, p. 77. Three men with devices to pump out flooded lands in Holland, (60, 30, 20). Schott. 1674. Ex. 1, pp. 570571. Cistern: (2, 3, 4, 1/4) done several ways. Cites Clavius for the lion fountain (Metrodorus 7). Wingate/Kersey. 1678?.  x  thQuest. 4, pp. 476477. Workmen: (20, 30). Quest. 5, pp. 477478. "I am a brazen lion ..." in Latin. (2, 3, 4, 1/2), where 6 hours is counted as 1/2 day, i.e. a day has 12 hours. = Metrodorus, Art. 7. Quest. 6, pp. 478479. (1/2, 10/7, 7/3). Quest. 7, pp. 479480. Dog, wolf, lion eating a sheep: (1, 3/4, 1/2), but the lion has a headstart of 1/8 hour. Quest. 12, p. 484. (20, x) in 12. Man and wife drinking beer. Quest. 13, pp. 484485. (30, 40, x) in 15. Carpenters building a house.  x  tW. Leybourn. Pleasure with Profit. 1694. Prob. 14, p. 39. Cistern emptying: (6, 4, 3, 2). Wells. 1698.  x  thNo. 105, p. 206. (20, 30) and (a, b). No. 106, p. 206. (20, x) in 12 and (a, x) in c.  x  tIsaac Newton. Arithmetica Universalis, 1707. ??NYS. English version: Universal Arithmetic, translated by Mr. Ralphson, with revisions and additions by Mr. Cunn, Colin Maclaurin, James Maguire and Theaker Wilder; W. Johnston, London, 1769. (De Morgan, in Rara, 652-653, says there were Latin editions of 1722, 1732, 1761 and Raphson's English editions of 1720 and 1728, ??NYS.) Resolution of Arithmetical Questions, Problem VII, pp. 184-185. "The Forces of several Agents being given, to determine x the Time, wherein they will jointly perform a given Effect d." Gives general approach for three workers. Example is (3, 8/3, 12/5), where the Force of the second is expressed as saying he can do the work "thrice in 8 weeks". Ozanam. 1725.  x  thProb. 24, question 9, 1725: 180-181. Prob. 5, 1778: 188189; 1803: 185186; 1814:161162; 1840: 84. Same as Metrodorus 7, except that a day is considered as 24 hours, so the problem is done as (2, 3, 4, ). Prob. 24, question 10, 1725: 181. (6, 9, 12) months to print a book.  x  tSimpson. Algebra. 1745. Section XI (misprinted IX in 1790).  x  thProb. XI, pp. 8384 (1790: prob. XXX, pp. 91-92). General problems, (a, b) and (a,b,c). Example: (10, 12, 16). Prob. XXXI, pp. 9092 (1790: prob. XXXVI, pp. 96-98). General problem: given (x,y) in a, (x, z) in b, (y, z) in c, determine x, y, z. Example with a,b,c=8, 9, 10.  x  tAlexisClaude Clairaut. El)mens d'Alg/bre. 1746. Vol. I, art. LVI (my source quotes from the 6th ed. of 1801). ??NX. He uses the context of this type of problem to study the meaning of negative solutions. A cistern of size a is filled by a source running for time b together with another source running for time c. Another reservoir of size d is filled by the sources in times e and f. Determine the rate of each source. Les Amusemens. 1749.  x  thProb. 173, p. 321. Cistern: (9, 12). Prob. 174, pp. 322323. Reservoir with three nymphs: (3, 4, 5).  x  t"By his Holiness the Pope". The Gentleman and Lady's Palladium (1750) 22. Qn. 11. (??NYS, cited by E. H. Neville; Gleaning 1259: On Gleaning 1146; MG 23 (No. 254) (May 1939) 149. "If a Cardinal can pray a soul out of purgatory ...." See Welch, 1833, below. Dictator Roffensis, proposer; Steph. Hodges & Will. Smith, solvers. Ladies' Diary, 175051 = T. Leybourn, II: 4546, quest. 334. [??NX of p. 46.] Three drinkers: (10, 12, 15) for 12 hour days, how long together for 10 hour days. Arthur Young. Rural Oeconomy: or, Essays on the Practical Parts of Husbandry. Dublin, 1770, p. 32. ??NYS described in: Keith Thomas; Children in early modern England; IN: Gillian Avery & Julia Briggs; Children and Their Books A Celebration of the Work of Iona and Peter Opie; OUP, (1989), PB ed, 1990, pp. 4577, esp. pp. 66 & 76. Proverb: one boy, one day's work; two boys, half a day's work; three boys, no work at all. Euler. Algebra. 1770. I.IV.III: Questions for practice.  x  thNo. 14, pp. 204-205. Cistern, (x, 20) in 12. No. 22, p. 205. Same as the example in Simpson's prob. XXXI  x  tVyse. Tutor's Guide. 1771?  x  thProb. 61, 1793: p. 69; 1799: p. 74 & Key p. 100. Two workers, (10, 13). Prob. 62, 1793: p. 69; 1799: pp. 7475 & Key p. 100. Boatbuilders, (18, x) in 11. Prob. 6, 1793: p. 128; 1799: p. 136 & Key p. 178. Cistern emptying, (1, 2, 3,). Gives volume of cistern but never uses it. = Metrodorus 133. Prob. 15, 1793: p. 156; 1799: p. 167 & Key p. 208. Builders, (34, x, 24) in 12. Prob. 5, 1793: p. 189; 1799: p. 201 & Key pp. 244245. Trenching a field. A, B, C can do in 12; B, C, D can do in 14; C, D, A can do in 15; D, A, B can do in 18. How long for all four together and for each one singly? Solution in decimals.  x  tDodson. Math. Repository. 1775.  x  thP. 22, Quest. LVIII. Man and wife drinking beer. (30, x) in 12. P. 23, Quest LIX. Cistern: (20, x) in 12. Pp. 5253, Quest. CV. Workers. A & B in 8; A & C in 9; B & C in 10. P. 56, Quest 56. (3, 8/3, 12/5), phrased as in Newton. Does the problem in general, then applies to the data.  x  tOzanamMontucla. 1778.  x  thQuestion 5, 1778: 193194; 1803: 190191; 1814: 165166; 1840: omitted. Same as Metrodorus 135. Prob. 22, 1778: 214; 1803: 209. Prob. 21, 1814: 181; 1840: 94. Same as the example in Simpson XXXI.  x  tCharles Hutton. A Complete Treatise on Practical Arithmetic and Bookkeeping. Op. cit. in 7.G.2. [c1780?] 1804: prob. 70, pp. 139140. Cistern: (40, 50, 25). Bonnycastle. Algebra. 1782.  x  thPp. 8283, no. 10 (1815: p. 102, no. 9). (10, 13). = Vyse, prob. 61. P. 83, no. 11 (1815: pp. 102103, no. 10). (a, b) done in general. P. 86, no. 25 (1815: p. 108, no. 34). Two drinkers: (30, x) in 12 days. P. 86, no. 26 (1815: p. 108, no. 36). Same as the example in Simpson's prob. XXXI. P. 86, no. 27. "If three agents, A, B, and C, can produce the effects a, b, c, in the times e, f, g, respectively; in what time would they jointly produce the effect d?"  x  tPike. Arithmetic. 1788.  x  thP. 335, no. 8. Cistern: (1/2, 1/4, 1/3). P. 350, no. 14. Merchant gaining and losing, equivalent to (7, 9, -2) how long to empty a full tank? P. 350, no. 19. Two workers: (7, 12). Pp. 350351, no. 20. Boatbuilders: (20, x) in 12. P. 351, no. 21. Two workers: (13, x) in 8. P. 351, no. 22. Three workers: (23, 37, x) in 15. P. 351, no. 23. Cistern: (55, 45, 30). P. 351, no. 24. Cistern of 73, inflow of 7/5 and outflow of 20/17 both run for two hours, then the outflow is stopped. P. 355, no. 39. A, B, C do a job. A and B do 3/11 of it, A and C do 5/13, Band C do 4/14. (Also entered at 7.G.1.)  x  tBullen. Op. cit. in 7.G.1. 1789. Chap. 38.  x  thProb. 34, p. 243. Mother & two daughters spin 3 lb flax in 1 day; mother can do it in 2 days; elder daughter in 2 days; how long does it take the younger daughter? Prob. 51, pp. 245-246. Cistern: (6, 8, 10, 12, -6, -5, -4, -3) how long to empty from full?  x  tEadon. Repository. 1794.  x  thP. 78, no. 21. If 3 men or 4 women can do a job in 68 days, how long will it take 2men and 3 women? P. 79, no. 26. If 5 oxen or 7 colts can eat a close in 87 days, how long will it take 2oxen and 3 colts? Answer is 105, which neglects growth of grass. P. 195, no. 10. (8, x) in 5. P. 195, no. 11. (32, 44, x) in 16. P. 195, no. 12. (3, 8/3, 12/5). = Newton's example. P. 367, no. 6. (40, 50, 25).  x  tJohn King, ed. John King 1795 Arithmetical Book. Published by the editor, who is the greatgreatgrandson of the 1795 writer, Twickenham, 1995. P. 100: (1, 2, 3); (7, 5, 6). Hutton. A Course of Mathematics. 1798?  x  thProb. 8, 1833: 212213; 1857: 216217. (6, 8), then generalises to (a, b, c, d). Prob. 15, 1833: 220221; 1857: 224225.. A and B can do in a days; A and C in b days; B and C in c days. How long for each singly and all three together? Prob. 39, 1833: 223; 1857: 227. (x, 30) in 12. Prob. 40, 1833: 223; 1857: 227. Problem 15 above with numerical values: a,b,c=8,9, 10.  x  tD. Adams. Scholar's Arithmetic. 1801. P. 125, nos. 26 & 27. (80, 40, 20, 10) & (27,15,12). Bonnycastle. Algebra. 10th ed., 1815. P. 226, no. 6. Cistern: (20, x) in 12 hours. Jackson. Rational Amusement. 1821. Curious Arithmetical Questions.  x  thNo. 4, pp. 15 & 71. A & B earn 40s in 6 days; A & C earn 54s in 9 days; B&C earn 80s in 15 days. What does each earn per day? No. 11, pp. 16 & 73. "A in five hours a sum can count, Which B can in eleven; How much more then is the amount They both can count in 7?" No. 29, pp. 21 & 8081. Lion, wolf, dog eating a sheep: (1/2, 3/4, 1), but the lion begins 1/8 before the others.  x  tSilvestre Fran'ois Lacroix. (l)mens d'Alg/bre, a l'Usage de l'(cole Centrale des QuatreNations. 14th ed., Bachelier, Paris, 1825. Section 14, ex. 1, pp. 2527. (5/2, 15/4) and (a, b) in general. Augustus De Morgan. Arithmetic and Algebra. (1831?). Reprinted as the second, separately paged, part of: Library of Useful Knowledge Mathematics I, Baldwin & Craddock, London, 1836. Arts. 3 & 112, pp. 12 & 2829. In Art. 3, he mentions the problem (10,16) as an example of algebraic formulation. In Art. 112, he solves it and (a, b). Welch. Improved American Arithmetic. 1833 ed. This must be Oliver Welch's American Arithmetic, first published in 1812 and which went through at least eight editions up to 1847. [Halwas 459465, of which 1833 is 462.] "If a Cardinal can pray a soul out of purgatory by himself in 1 hour, a bishop in 3 hours, a Priest in 5 hours, a Friar in 7 hours, in what time can they pray out 3 souls, all praying together?" In the 1842 ed., this was changed to steam, water, wind and horse power. ??NYS quoted in Gleaning 1146, MG 21 (No. 245) (Oct 1937), 258. See above at 1750 for an earlier version. Nuts to Crack II (1833), no. 129. (30, x) in 12. Identical to Bonnycastle, 1782, no. 25. Bourdon. Alg/bre. 7th ed., 1834. Art. 57, p.85. General solution for (a/b, c/d, e/f). D. Adams. New Arithmetic. 1835.  x  thP. 243, no. 74. Ship has a leak which will fill it in 10 and a pump which will empty it in 15, i.e. (10, 15). P. 247, no. 109. Two workers. (3, 4). P. 247, no. 110. Three workers. A & B can do in 4; B & C in 6; A & C in 5. P. 247, no. 111. Two workers. (7, x) in 5.  x  tAugustus De Morgan. On The Study and Difficulties of Mathematics. First, separately paged, part of: Library of Useful Knowledge Mathematics I, Baldwin & Craddock, London, 1836. P. 31. (12, 15), then does (a, b). Augustus De Morgan. Examples of the Processes of Arithmetic and Algebra. Third, separately paged, part of: Library of Useful Knowledge Mathematics I, Baldwin & Craddock, London, 1836.  x  thProb. 16, p. 28. Given (5, 7), how long will it take to do A? P. 91. A & B in c; A & C in b; B & C in a.  x  tUnger. Arithmetische Unterhaltungen. 1838.  x  thPp. 135 & 258, no. 519. A can do 63 in 8 days; B can do 37 in 6 days and C can do 25 in 3 days. How long for all three to do 268? Pp. 135 & 258, no. 520. Cistern (20, 15, 12). Pp. 135136 & 258, no. 521. Cistern (12, 8, 10).  x  tHuttonRutherford. A Course of Mathematics. 1841? Prob. 8, 1857: 81. Two workers: (12,14). T. Tate. Algebra Made Easy. Op. cit. in 6.BF.3. 1848.  x  thP. 75, no. 11. (x + 6, x + 3/2) in x. P. 76, no. 12. (x, x+5) in 6. P. 85, no. 8. A can reap a field in a days. If assisted by B for b days, then A only has to work c days.  x  tFamily Friend 1 (1849) Answers to correspondents, pp. 4 & 6. Questions requiring answers. No. 1. (12, x) in 3. Anonymous. A Treatise on Arithmetic in Theory and Practice: for the use of The Irish National Schools. Third edition, revised, improved and enlarged. Published by direction of the Commissioners of National Education in Ireland, Dublin, 1850. Many examples, of which the following are the more interesting.  x  thPp. 200201, no. 133. (17/2, 21/4, x) in 6/5. P. 347, no. 8. (15, x) in 10. P. 347, no. 11. 5 mills grind 7, 5, 4, 3, 1 bushels per hour. How long for all five to grind 500 bushels? P. 356, no. 10. A, B & C can do in 10 days; B & C can do in 16 days; how long for A? (This is equivalent to (x, 16) in 10.) P. 358, no. 30. A, with 2 days' help from B, can do in 12 days. B, with 4 days' help from A, can do in 8 days. How long for both together? P. 358, no. 31. A and B can do in 8 12hour days. A can do in 12 16hour days. How many 14hour days would B need? P. 359, no. 36. (5/2, 9/4, x) in 1. P. 359, no. 39. (6, 8, 10, 12, 6, 5, 4, 3) how long to empty from full?  x  tJohn William Colenso (18141883). Arithmetic Designed for the Use of Schools .... New edition. Longman, Brown, Green, and Longmans, 1853. ??NX Wallis 246 COL. I have 1857, which seems identical for pages 1 164, then adds a chapter on decimal coinage on pp. 165171. I also have 1871, which rearranges the material at the end and adds Notes and ExaminationPapers the advertisement on p. v says the additional material was added by J. Hunter in 1864. This gives a large number of variations of the problem which I have included here as representative of mid 19C texts. Miscellaneous Examples, pp. 122-136, with answers on pp. 161163 (1871: 211-213).  x  thNo. 21. (10, 13). No. 27. (10, B) in 7. No. 31. Cistern, (40, 50, -25). No. 52. (3, 8/3, 12/5), where B is determined from "B can do thrice as much in 8 days". I.e. Newton's example; see also Eadon. No. 80. If 5 oxen or 7 horses can eat the grass of a field in 87 days, how long will it take 2 oxen and 3 horses? (The grass is not assumed to grow.) = Eadon, No.26. No. 101. If 3 men, 5 women or 8 children can do a job in 26 hours, how long will it take 2 men, 3 women and 4 children?  x  t ExaminationPaper VIII, (1864), 1871: pp. 170-172, with answers on p. 214.  x  thNo. 1. M can do in 20 7-hour days. N can do in 14 8-hour days. How many hours per day must they work together to do it in 10 days? No. 2. Cistern, (20, 24, -30). How full is it after 15? No. 3. Reapers, (F, G) in 8, with 3 : F = 5 : G. No. 4. (34, 38), but second man stops 4 days before the end. No. 5. Cistern. (A, B) fills in 4; (A, -C) empties in 40; (B, -C) fills in 60. No. 6. 4 men, working various parts of the time. No. 7. (A, B) in 14, (B, C) in 10, (A, C) in 12. No. 8. B = twice (A, C); C = thrice (A, B); (A, B, C) in 5. No. 9. Three men working various parts of the time. No. 10. Cistern with 2 inlets and 2 outlets running various parts of the time.  x  tJohn Radford Young (17991885). Simple Arithmetic, Algebra, and the Elements of Euclid. IN: The Mathematical Sciences, Orr's Circle of the Sciences series, Orr & Co., London, 1854. [An apparently earlier version appeared as Circle of the Sciences, vol. (2 &) 3, with no TP or other details except two editorial remarks referring to Professor Young as author of the Arithmetic and Algebra sections. I have vol. 3, which begins with the last sheet of Arithmetic and then covers the other material. (I saw a vol. 8? on zoology? in the same bookshop.) This version includes detailed solutions and some plates not connected with the text. The 1854 text is identical except that the order of topics has been changed and there are some consequent changes to the text.] This is a typical mid 19C text with a number of cistern problems, of which the more interesting are the following.  x  thNo. 16, p. 178. A can do a job in 10 days. After he has worked for 4 days, B comes to assist and they finish it in 2 more days. How long would B take by himself? No. 4, p. 207. A man and his wife can drink a barrel in 15 days. After 6 days the man leaves and the woman finishes it in 30 days. How long would it take her to drink the whole barrel by herself? No. 10, p. 208. Similar to the last, with two workers and numbers 16, 4, 36. No. 12, p. 208. General solution for (a, b, c).  x  tGerardus Beekman Docharty. A Practical and Commercial Arithmetic: .... Harper & Brothers, NY, 1854. Many examples on pp. 166167, 242243, 247, including Simpson's XXXI; the same problem with values 12, 20, 15; (35/2, x) in 40 and the following.  x  thP. 167, no. 36. A, B and C can do in 24; A and B can do in 48; A and C can do in 36. How long for each separately? Pp. 247248, no. 64. = Vyse, prob. 5, but only asks how long for all together, and gives solution in fractions.  x  tVinot. 1860. Art. LVII: Les trois Ouvriers, pp. 7475. Same as the example in Simpson XXXI. James B. Thomson. Higher Arithmetic; or the Science and Application of Numbers; .... Designed for Advanced Classes in Schools and Academies. 120th ed., Ivison, Phinney & Co, New York, (and nine copublishers), 1862. Lots of straightforward examples and the following.  x  thProb. 65, p. 397 & 422. (15, x) in 18. Note that x has a negative value, i.e. is an outlet. See also BR & Docharty. Prob. 93, p. 398 & 422. = Vyse, prob. 5, with solution in fractions.  x  tEdward Brooks. The Normal Mental Arithmetic A Thorough and Complete Course by Analysis and Induction. Sower, Potts & Co., Philadelphia, (1858), revised, 1863. Further revised as: The New Normal Mental Arithmetic: A Thorough and Complete Course by Analysis and Induction; Sower, Potts & Co., Philadelphia, 1873. Lots of examples. I mention only those of some novelty to illustrate mid/late 19C texts.  x  th1863 p. 122; 1873 pp. 140141; no. 16. (6, 8, x) in 3. 1863 p. 122; 1873 p. 141; no. 18. (A, B, C) in 4; (A, B) in 6; (B, C) in 9. 1863 pp. 122123; 1873 p. 141; no. 19. (A, B, C) in 6; (A, B) in 8; B in 12. 1863 p. 123; 1873 p. 141; no. 26. (A, B, C) can do in 6; (A, B) can do in 9; all three work for 2 days, then C leaves how long for A, B to finish? 1863 p. 155, no. 2; 1873 p. 174, no. 9. (A, B, C) can do in 20; (A, B) can do in 30; (B, C) can do in 40; all three work for 5 days, then B leaves how long for A, C to finish?  x  tBoy's Own Magazine 2:2 (No. 8) (Aug 1863) 183 & 2:4 (No. 11) (Nov 1863) 367. (Reprinted as (Beeton's) Boy's Own Magazine 3:8 (Aug 1889) 351 & 3:10 (Oct 1889) 431.) Mathematical question 87. Complicated version, typical of its time. Bacchus drinks from a cask for A of the time it would take Silenus to drink the whole cask. Silenus then finishes it off and the total time is two hours longer than if they had drunk together. But if they had drunk together, Bacchus would only have drunk half as much as he left for Silenus. [Robert Chambers]. Arithmetic. Theoretical and Practical. New Edition. Part of: Chambers's Educational Course edited by W. & R. Chambers. William and Robert Chambers, London and Edinburgh, nd, [1870 written on fep]. [Though there is no author given, Wallis 242 CHA is the same item, attributed to Robert Chambers, with 1866 on the flyleaf, so I will date this as 1866? ??check in BMC.]  x  thP. 263, quest. 4. Person doing business, equivalent to a cistern of 8000 with inlets of 1500 and 1000 per year and a drain of 3000 per year. When is he broke? This is not really a cistern problem since the rates are given rather than the times to fill or empty, but the format is sufficiently similar that I have included it here as an example of the more straightforward rate problems. Also, the formulation with money is not common. P. 266, quest. 44. Cistern (10, 8), but the second pipe is not turned on until the cistern is half full.  x  tStoddard, John F. The American Intellectual Arithmetic: Containing An Extensive Collection of Practical Questions on the General Principles of Arithmetic. With Concise and Original Methods of Solution, Which Simplify Many of the Most Important Rules of Arithmetic. Sheldon & Company, New York & Chicago, 1866. P. 120, no. 30. "If a wolf can eat a sheep in E of an hour, and a bear can eat it in  of an hour, how long would it take them together to eat what remained of a sheep after the wolf had been eating  of an hour?" Thanks to David E. Kullman for sending this. Todhunter. Algebra, 5th ed. 1870. Many examples the less straightforward are the following.  x  thExamples X, no. 33, pp. 87 & 577. (C/3, 2C/3, C) in 6. Examples XXIV, no. 23, pp. 212 & 586. (A, A2) in 15/8. Miscellaneous Examples, no. 48, pp. 548 & 604. Simpson's XXXI with values 12,15,20. Cf Docharty.  x  tBachetLabosne. Problemes. 3rd ed., 1874. Supp. prob. V, 1884: 187188. Poetic and complicated form. In 3/10 of the time that Silenus would take to drink the whole amphora, Bacchus drinks 1/4 of what he leaves for Silenus to finish. But if they drank it all together, they would finish it in two hours less than the previous time. Daniel W. Fish, ed. The Progressive Higher Arithmetic, for Schools, Academies, and Mercantile Colleges. Forming a Complete Treatise of Arithmetical Science, and its Commercial and Business Applications. Ivison, Blakeman, Taylor & Co., NY, nd [but prefaces give: 1860; Improved Edition, 1875]. P. 419, no. 80. Five persons building a house. All but A can do in 14 days; all but B can do in 19; all but C can do in 12; all but D can do in 15; all but E can do in 13. How long for all five and who is the fastest worker? Answer: 11 4813/12137. M. Ph. Andr). (l)ments d'Arithm)tique (No 3) a l'usage de toutes les institutions .... 3rd ed., Librairie Classique de F.E. Andr)Gu)don, Paris, 1876. Several straightforward problems and the following. Prob. 99, p. 62. Four companies of workers can do a job in 45, 9, 27, 36 days. How long will it take 2/5 of the first company, 3/4 of the second company, 1/2 of the third company and 1/3 of the fourth company? Fred Burnaby. On Horseback Through Asia Minor. Sampson Low, et al., London, 1877. Vol. 1, pp. 208-210. One man can mow in 3 days, the other in 4. How long together? Mittenzwey. 1880.  x  thProb. 7073, pp. 14 & 6566; 1895?: 7780, pp. 1819 & 68; 1917: 7780, pp. 1718 & 6465. Cistern problems: (6, 8); (x, 20) in 12; (9, 6, 4); same with openings delayed by 2, 1, 3 hours. His solutions indicate the general method. Prob. 81, p. 1516 & 6667; 1895?: 89, pp. 20 & 69; 1917: 89, pp. 1819 & 66. Three drinkers, (12, 10, 8).  x  tWilliam J. Milne. The Inductive Algebra .... 1881. Op. cit. in 7.E.  x  thNo. 106, pp. 142 & 332. Asks for general solution of (a, b) and two specific cases. No. 6, pp. 162 & 334. (8, 9, 10). No. 2, pp. 166 & 334. General solution of (a, b) and one specific case.  x  tWilliam Briggs & George Hartley Bryan. The Tutorial Algebra, based on the Algebra of Radhakrishnan Part II Advanced Course. W. B. Clive, London, (1898), 1900. Exercises X, prob. 6, pp. 123 & 579. Two reapers, (x, x-5) in 6. A. Sonnenschein & H. A. Nesbit. The New Science and Art of Arithmetic For the Use of Schools. A. & C. Black, London, 1903. This is a typical text of its time and has a number of variations on the basic problem.  x  thPp. 233234 & 479, prob. 37. Workers (10, 12, 9). How long will it take to do 2 tasks how much does each person do? Pp. 234 & 479, prob. 38. Workers: man, woman & boy can do in (5, 8, 12). Man works 1 days, then is joined by the woman for 1 days, then the remainder is left to the boy. When does he finish and how much does each do? Pp. 234 & 479, prob. 44. Cistern of size 600 (5, 7). How much does each pipe pass? Pp. 234 & 479, prob. 45. Workers (A, B) in 6; (A, C) in 8; (B, C) in 9. Pp. 235 & 479, prob. 54. Cistern (4, 6, 4). Pp. 288 & 484, prob. 42. Two pumps and they can work at half or full speed. If (A,B/2) in 5 and (A/2, B) in 4, determine A, B and (A,B).  x  tPearson. 1907. Part II, no. 162, pp. 145 & 223. A brings a pint every 3 minutes, B a quart every 5 minutes and C a gallon every 7 minutes. How long to fill a 55 gallon drum and who finishes the job? Stephen Leacock. A, B, and C. IN: Literary Lapses, (1910). The book has been frequently reprinted and the piece has been widely anthologised. It is pp. 237245 in my 9th English ed. Collins. Fun with Figures. 1928. According to Hoyle, not arithmetic, pp. 3233. "[I]f your father can build a chicken coop in 7 days and your Uncle George can build it in 9 days, how long ...." "They'd never get it done; they'd sit down and swap stories of rum runners, and bootleggers and hijackers." C. Dudley Langford. Note 1558: A graphical method of solving problems on "Rate of Work" and similar problems. MG 25 (No. 267) (Dec 1941) 304307. + Note 2110: Addition to Note 1558: "Rate of Work" problems. MG 34 (No. 307) (Feb 1950) 44. Uses a graph to show (a,b) problems as meeting problems. Also solves problems (A,x) in B and (a,b), the latter appearing as an overtaking problem. The Addition gives a clearer way of viewing (a,-b) problems as overtaking problems. David Singmaster. How long is a brick wall? (my title is: Three bricklayers). Weekend Telegraph (30 May 1992) xxxii & (7 Jun 1992) xxx. Al and Bill can build a wall in 12 days; Al & Charlie in 15; Bill & Charlie in 20. How long does it take each individually and how long does it take all three together? This is well known, but then I ask how can you determine integer data to make all the results come out integers? Let A, B, C denote the amounts each can build in a day. To make all the data and results come out as integers, we have to have all of A, B, C, A+B, A+C, B+C, A+B+C be fractions with unit numerators. To combine them easily, imagine that all these fractions have been given a common denominator d, so we can consider A = a/d, B = b/d, etc., and we want a, b, c, a+b, a+c, b+c, a+b+c to all divide d. We can achieve this easily by taking any three integers a, b, c, and letting d be the least common multiple of a, b, c, a+b, a+c, b+c, a+b+c. Taking a, b, c = 3, 2, 1, we find d=60 and the given problem is by far the simplest example with distinct rates A, B, C. I feel this is based on my remembering the problem from somewhere, but the only previous use of this data is in Docharty, but he doesn't consider the diophantine problem, and none of the other data has this property. I suspect this was an AMM or similar problem some years ago. John Silvester recently asked me if I knew the following version, which he heard from John Reeve. A man can pack his bag to go to a meeting in 20 minutes. But if his wife helps him, it takes an hour. How long would it take his wife on her own? I.e. (20, x) in 60. The answer is 30 minutes! 7.H.1.WITH GROWTH NEWTON'S CATTLE PROBLEM  t The example of Ray has led me to reexamine the relation between this topic and the cistern problems. Ray's problem can be recast as follows. There is a cistern with an input pipe and a number of equal outlet taps. When a taps are turned on, the cistern empties in time c; when d taps are turned on, the cistern empties in time f; how long [h] will it take to empty when x taps are turned on? Despite the similarity, we do not have the times for the individual inlets and outlets to fill or empty the cistern and so it is much easier to use rates rather than times.  t Isaac Newton. Arithmetica Universalis, 1707. ??NYS. English version: Universal Arithmetic, translated by Mr. Ralphson, with revisions and additions by Mr. Cunn, Colin Maclaurin, James Maguire and Theaker Wilder; W. Johnston, London, 1769. (De Morgan, in Rara, 652-653, says there were Latin editions of 1722, 1732, 1761 and Raphson's English editions of 1720 and 1728, ??NYS.) Resolution of Arithmetical Questions, Problem 11, pp. 189-191. (Sanford 165 quotes from the 1728 English edition and it is the same as the 1769.) "If the Number of Oxen a eat up the Meadow b in the time c; and the Number of Oxen d eat up as good a Piece of Pasture e in the Time f, and the Grass grows uniformly; to find how many Oxen [x] will eat up the like Pasture g in the Time h." Gives a general solution: (gbdfh-ecagh-bdcgf+ecfga)/(befh bceh) and an example with a,b,c;d,e,f;g,h=12,3@,4; 21, 10, 9; 24, 18. One easily finds that the rate of grass growth per unit area per unit time is G = (ace bdf)/cf(bd ae) and the rate of grass consumption per ox per unit time is E = be(cf)/cf(bd ae). There are actually three unknowns since we also don't know the initial amount of grass per unit area, G0. We can either take proportions of these or we can adopt a unit of grass such that the initial amount of grass per unit area is 1. In the second case, the basic equation b(G0Ġ+Gc)=aEc becomes b(1+ cG) = acE. Walkingame. Tutor's Assistant. 1751. 1777: p. 177, prob. 121; 1860: p. 185, prob. 114. Identical to Newton. Eadon. Repository. 1794.  x  thPp. 208209, no. 6. Same as Newton. Gives a specific solution. "This is deemed a curious and difficult question, it was first proposed by Sir Isaac Newton, in his Universal Arithmetic, and there universally solved by an algebraic process: but I have never seen a numerical solution independent of algebra, except this of my own; ...." He then states the general solution as [bdfg(hc) + aceg(fh)]/beh(fc). Pp. 209210, no. 7. "If 3 oxen or 5 colts can eat up 4 1/5 acres of pasture in 7 weeks, and 5 oxen and 3 colts can eat up 9 acres of like pasture in 10 weeks, the grass gowing [sic] uniformly; how many sheep will eat up 48 acres in 20 weeks, supposing 567 sheep to eat just as much as 6 oxen and 11 colts?" Answer is 1736. He introduces heifers, where a heifer is 1/5 of an ox or 1/3 of a colt, which brings the problem into Newton's form with values a,b,c;d,e,f;g,h=15, 4 1/5, 7; 34, 9, 10; 48, 20, which gives the answer in terms of heifers, which he converts to sheep.  x  tStoddard, John F. Stoddard's Practical Arithmetic. The Practical Arithmetic, Designed for the use of Schools and Academies; Embracing Every Variety of Practical Questions Appropriate to Written Arithmetic, with Original, Concise and Analytic Methods of Solution. Sheldon & Company, NY, 1852. Pp. 281282, no. 23. a,b,c;d,e,f;g,h=14, 2, 3; 16, 6, 9; 24, 6. Thanks to David E. Kullman for sending this. Parlour Pastime, 1857. = Indoor & Outdoor, c1859, Part 1. = Parlour Pastimes, 1868. Arithmetical puzzles, no. 6, p. 174 (1868: 185): Sir Isaac Newton's problem. Gives the numerical values from Newton, but 3@ is often misset as 3, and when so done, is identical in all three editions. Vinot. 1860. Art. LI: Les boeufs de Newton, pp. 6768. a,b,c;d,e,f;g,h=3,2,2;2,2,4; 6, 6. Answer: 5. A. Schuyler. A Complete Algebra for Schools and Colleges. Van Antwerp, Bragg & Co., Cincinnatti & NY. Pp. 99100, nos. 3132, gives Newton's specific problem, then the general version. Thanks to David E. Kullman for sending this. M. Ph. Andr). (l)ments d'Arithm)tique (No 3) a l'usage de toutes les institutions .... 3rd ed., Librairie Classique de F.E. Andr)Gu)don, Paris, 1876. Same as Vinot, but no answer. Joseph Ray, revised by J. M. Greenwood. Ray's New Higher Arithmetic. American Book Co., Cincinnati, 1880. ??NYS quoted and discussed in: David E. Kullman; Story problems with a flavor of the old Northwest; preprint of 8pp sent by the author in Apr 1999, p. 4. "There is coal now on the dock, and coal is running on also, from a shoot [sic], at a uniform rate. Six men can clear the dock in one hour, but 11 men can clear it in 20 minutes: how long would it take 4 men? Ans: 5 hr." This is like Newton's problem, but assuming b = e = g, and solving for h rather than x. In a letter Kullman notes that this problem is not in Ray's original edition, called Higher Arithmetic, date not given, but Ray died in 1855. G. H. Mapleton, proposer; Charles Hammond, solver. Arithmetical problem. Knowledge 1 (30 Dec 1881) 191 & (20 Jan 1882) 258, item 9. a,b,c;d,e,f;g,h=12,10,16;18, 10, 8; 40, 6. Cites Newton, 1722 ed., p. 90. Charles Pendlebury. Arithmetic. Bell, London, (1886), 30th corrected and expanded printing, 1924. Section XLIV: Pasture with growing grass, pp. 336q336s & answers, part II, p.xv. (These pages were added after the 6th ed. of 1893, by the 10th ed. of 1897, possibly for the 10th ed?) This section carefully works an example to answer different questions, then gives 12 problems. In all cases, all the fields have the same size, which much simplifies things.  x  thThe example problem has a,b,c;d,e,f;g,h=70,b,24;30,b,60; b, 96. Prob. 12. A cistern has a number of equal holes in its base and a pipe is adding water. When 10 holes are open, the cistern will empty in 20 minutes. When 8 are open, it empties in 35 minutes. How long will it take with 12 holes open? This is equivalent to a,b,c;d,e,f;g,h=10,b,20;8,b,35; b, 12.  x  tI have read an article (in HM??) which described this as a common problem in 19C textbooks. Walter Percy Workman. The Tutorial Arithmetic. University Tutorial Press, (1902); 2nd ed., 1902. [There is a 3rd ed., (c1908), c1928, which contains a few more pages.] SectionIX [= Chap. XXXI in the 1928 ed.], examples CXLV, prob. 59-60, pp. 430 & 544 [= 436 & 577 in the 1928 ed.].  x  thProb. 59. a, b, c; d, e, f; g, h = 15, b, 8; 9, b, 16; b, 12 this has all the fields being the same. Prob. 60 is more complex. The field is divided into two equal halves. 50 oxen eat all of one half in 4 weeks. Four oxen are slaughtered and the rest are put in the other half. After 6 weeks, 3 oxen are slaughtered. In another week, all the grass is eaten up and the oxen are sold. After another week, the division is removed and 85 oxen are put in the field. How long will it last them?  x  tWehman. New Book of 200 Puzzles. 1908. An ox problem, p. 55. Gives Newton's problem with just the numerical values. Loyd. Cyclopedia, 1914, pp. 47 & 345. = MPSL1, prob. 48, pp. 46 & 138-139. Cow, goat and goose. Wood. Oddities. 1927. Prob. 71: Ox and grass, p. 55. a,b,c;d,e,f;g,h=6,10,16;18,10, 8; 40, 6. Gets 88, but is confused about the growing of the grass Newton's formula gives 104. Says Newton divides the oxen into those that eat the accumulated grass and those that eat the increase, but he doesn't apply this correctly. Indeed, for this data, the grass grows at a negative rate! Undoubtedly intended to be the data of Mapleton, 1881, for which the answer is 88. Perelman. MCBF. 1937.  x  thCows in the meadow. Prob. 139, pp. 229234. Same as Pendlebury's example. Newton's problem. Prob. 140, pp. 234235. Same data as Newton.  x  tA. I. Ostrovsky. Oxen grazing in a field. MG 50 (No. 371) (Feb 1966) 46-48. Quotes Newton and gives a graphical solution which converts this into an overtaking problem, where the grass starts growing 12 weeks before the cows are put in. John Bull. Grazing Oxen. M500 165 (Dec 1998) 14. He is unhappy with some of the limiting situations and proposes a different basic equation. However, his unhappiness is really due to not understanding the basic equation properly.  7.H.2.DIVISION OF CASKS  tNOTATION: (a, b, c) among n means to divide a full, b half-full and c empty casks among n people so that each has an equal amount of contents and of casks. Dividing kn casks containing 1, 2, ..., kn among k people so each gets the same amount of contents and of casks see: Albert; Munich 14684; AR; GGnther (1887); Singmaster (1998). See Tropfke 659.  tAlcuin. 9C.  x  thProb. 12: Propositio de quodam patrefamilias et tribus filius ejus. (10, 10, 10) among 3. This has 5 solutions he gives just 1: 0, 10, 0; 5, 0, 5; 5, 0, 5. Prob. 51: Propositio de vino in vasculis a quodam patre distributo. Divide 4 casks containing 10, 20, 30, 40 among 4 solution involves shifting contents, so this is not really a problem of the present type.  x  tAbbot Albert. c1240. Prob. 3, p. 333. Divide 9 casks containing 1, 2, ..., 9 among three. He gives only one of the two solutions: 1, 5, 9; 2, 6, 7; 3, 4, 8. See Singmaster, 1998, for a generalization. BR. c1305. No. 40, pp. 58-61. 300 ewes, 100 each with 1, 2, 3 lambs, to be divided among three sons so each son has the same number of ewes and lambs and no lamb is separated from its mother. This is the same as (100,100,100) among 3. He gives one solution: 0, 100, 0; 50, 0, 50; 50, 0, 50. [There are 234 solutions!] Munich 14684. 14C. Prob. XXIV, f. 32r. Same as Abbot Albert and with the same solution. Folkerts. Aufgabensammlungen. 1315C. 10 sources of Abbot Albert's problem. Also cites Albert, AR, GGnther. AR. c1450. Prob. 351, pp. 154, 182. Same as Abbot Albert, with the same solution, but arranged in columns. Vogel comments that this makes a 'half-magic square' and cites GGnther, 1887, as having already noted this. Tartaglia. General Trattato, 1556, art. 130-131, p. 255v.  x  thArt. 130: (7, 7, 7) among 3. Gives one of the two solutions: 3, 1, 3; 3, 1, 3; 1, 5, 1. Art. 131: (9, 9, 9) among 3. Gives one of the three solutions: 4, 1, 4; 3, 3, 3; 2, 5, 2.  x  tBachet. Problemes. 1612. Addl. prob. IX: Trois hommes ont ! partager 21 tonneaux ...., 1612:161-164; 1624: 233236; 1884: 168-171.  x  th(7, 7, 7) among 3 gives all two solutions. 1612 cites Tartaglia. (9, 9, 9) among 3 gives all three solutions. 1612 cites Tartaglia. (8, 8, 8) among 4 Bachet erroneously does (6, 12, 6) among 4, but the editor gives all four solutions of the original problem. Labosne adds: (5, 11, 8) among 3 giving all three solutions. (Ahrens, A&N, 29, says that the 1st ed. also does (5, 5, 5) among 3.)  x  tvan Etten. 1624. Prob. 89 (86), part IV, pp. 134-135 (213). (7, 7, 7) among 3. One solution: 3, 1, 3; 3, 1, 3; 1, 5, 1. Hunt. 1651. Pp. 284285. Of three men that bought wine. (7, 7, 7) among 3. Two answers. Ozanam. 1725. Prob. 44, 1725: 242-246. Prob. 24, 1778: 182184; 1803: 180182; 1803:158159; 1814: 158159; 1840: 8182.  x  th(7, 7, 7) among 3 gives both solutions. Notes that this cannot be divided among 4 persons because 4 does not divide 21. 1725 has a very confusing attempt at (8, 8, 8) among 4, which is done as though it were (6, 12, 6) among 4, and he seems to think half-empty is different than half-full!! 1778 onwards just do (8, 8, 8) among 3, giving 3 of the 4 solutions, omitting 4, 0, 4; 4, 0, 4; 0, 8, 0. (9, 9, 9) among 3 gives all 3 solutions.  x  tLes Amusemens. 1749. Prob. 15, p. 137: Les Tonneaux.  x  th(7, 7, 7) among 3 gives both solutions. (11, 11, 11) among 3 gives 2 of the 4 solutions.  x  tBestelmeier. 1801. Item 717: Die sonderbare Weintheilung unter 3 Erben. Says there are 21 casks, so presumably (7, 7, 7) among 3. Jackson. Rational Amusement. 1821. Arithmetical Puzzles.  x  thNo. 11, pp. 3 & 53. (7, 7, 7) among 3. Both solutions. No. 51, pp. 11 & 66. (8, 8, 8) among 3. Gets three of the four solutions, omitting 4,0,4; 4,0, 4; 0, 8, 0.  x  tEndless Amusement II. 1826? Prob. 20, pp. 199200. (7, 7, 7) two solutions. Young Man's Book. 1839. Pp. 239240. Identical to Endless Amusement II. Walter Taylor. The Indian Juvenile Arithmetic .... Op. cit. in 5.B. 1849. P. 211. (7, 7, 7) with both solutions. Magician's Own Book. 1857. The wine and the tables, p. 225. (7, 7, 7), (8, 8, 8), (9, 9, 9) among 3 gives two solutions for each. = Boy's Own Conjuring Book, 1860, p. 195. Vinot. 1860. Art. XL: Un partage curieux, pp. 5960. (7, 7, 7) two solutions. Leske. Illustriertes Spielbuch fGr Mdchen. 1864? Prob. 593, pp. 299 & 410: Sechs KnacknGsse part 1. (7, 7, 7) one solution shown diagrammatically: 3,1,3;3,1,3;1, 5, 1. Mittenzwey. 1880. Prob. 104, pp. 21 & 73; 1895?: 121, pp. 26 & 75; 1917: 121, pp. 24 & 73. (7, 7, 7) among 3. Gives three solutions, but one is a rearrangement of another. Solution asks, if pouring is allowed, can all three get the same inheritance? It says to pour two half barrels into two other half barrels, obtaining (9, 3, 9). Siegmund GGnther. Geschichte des mathematischen Unterrichts im deutschen Mittelalter bis zum Jahre 1525. Monumenta Germaniae Paedagogica III. 1887. Facsimile reprint by Sndig Reprint Verlag, Vaduz, Liechtenstein, 1969. He discusses Abbot Albert and his problem on pp. 3536, noting that the solution can be viewed as a set of lines in a magic square so that the perpendicular lines give a second solution, but that magic squares were then unknown in Europe. He gives no other examples. Lucas. L'Arithm)tique Amusante. 1895. Prob. XV: Jeux de tonneau, pp. 5051. (7, 7, 7) among 3. Gives both solutions. Notes that half empty = half full, so doubling gives us empty = full! Ahrens. A&N. 1918. Pp. 29-33. Gives all solutions for (n, n, n) among 3 for n = 5 (1) 10; (8, 8, 8) among 4 & 6; (11, 5, 8) among 3; (5, 11, 8) among 3; (4, 12, 8) among 3. McKay. Party Night. 1940. No. 14, p. 178. (7, 7, 7) among three. One solution. M. Kraitchik. Mathematical Recreations, 1943, op. cit. in 4.A.2, chap.2, prob. 34, pp. 31-32. 9 full, 9 threequarter full, 9 half, 9 quarter full, 9 empty among 5. David Singmaster. Triangles with integer sides and sharing barrels. CMJ 21 (1990) 278-285. Shows the number of ways of sharing (N, N, N) among 3 is the same as the number of integer sided triangles of perimeter N. Shows this is the number of partitions of N/2 or (N-3)/2 into 3 parts. Finds necessary and sufficient conditions for sharing (a,b, c) among k people. David Singmaster. Fair division of the first kn integers into k parts. Written in 1998. This generalizes the problem of Abbot Albert and determines that there is a partition of the first kn integers into k sets of n values with each set having the same sum if and only if n is even or n > 1 and k is odd. I call this a fair division of the first kn integers into k parts. The number of such divisions is large, but might be worth examining. 7.H.3.SHARING UNEQUAL RESOURCES PROBLEM OF THE PANDECTS  tNOTATION: (a, b, ...; c) means persons contribute a, b, ... which is shared equally among themselves and an extra person who pays c. See Clark for a variant formulation with the same result. See McKay; Party Night; 1940 in 7.H.5 for a related form. INDEX of (a, b, ...; c) problems with a  b  .... x4 "  H2H3H3Kraitchik H2H3H5Fibonacci; alQazw3n3; Bartoli; Calandri H2H3H35McKay H3H4H4Kraitchik H3H4H7Kraitchik H3H4H14Gherardi H3H5H4Vinot H3H5H5Benedetto da Firenze H3H5H8D. Adams, 1801; Jackson; Badcock; New Sphinx; Magician'sOwnBook; Bachet-Labosne; Mr.X; Pearson; Clark; Kraitchik; Rohrbough; Sullivan H3H5H10AR H3H5H80Kraitchik H4H5H4Pseudodell'Abbaco H4H6H10Mittenzwey H5H7H12Kraitchik H7H8H30Kraitchik H10H14H6Hummerston H31H50H40Tagliente H90H120H70Kraitchik x4 <"" H11H14H17H42Kraitchik x4 <""  H2H3H6H9H4Kraitchik  t Fibonacci. 1202. P. 283 (S: 403404). (3, 2; 5). Qazwini = Zakariy ibn Muhammad ibn MahmEd abE Yahya alQazw3n3. (= alKazw3n3 = Zakariyy]' b. Muhammad b. Mahmd Ab Yahya alKazwn = Zakary] ibn Muhammad alQazwn). (Kitb) Ajib al-MakhlEqt wa Gharib alMawjEdt (= Adj]yib alMakhlk]t wa Gh]r]'ib alMawdjd]t = Aj]ib al-makhlq]t waghar]ib almawjd]t) ((The Book of the) Wonders of the Creation and Unique [Phenomena] of the Existence = Prodigies of Things Created and Miraculous Aspects of Things Existing = The Wonders of Creation and the Peculiarities of Existing Things = The Cosmography). c1260. ??NYS the earliest dated copy, of 1458, and several others are in the Wellcome Institute; BL has a page from a 14C copy on display. Part 8: On the arts; chap. 9: On reckoning. In: J. Ruska; Kazwnstudien; Der Islam 4 (1913) 14-66 & 236-262. German translation (Arabic omitted) of this problem on pp. 252-253. (3,2;5). Story says one proposes a 3 : 2 split, but 4 : 1 is found to be correct. [Qazwini also wrote a Geography, in two editions, and its titles are slightly similar to the above. I previously had reference to the Arabic titles of the other book, but rereading of Ruska and reference to the DSB article shows the above is correct.] Gherardi. Libro di ragioni. 1328. Pp. 40-41: Chopagnia. (3, 4; 14). Bartoli. Memoriale. c1420. Prob. 26, ff. 77v78r (= Sesiano, pp. 144 & 149). (2, 3; 5), correctly solved. Pseudodell'Abbaco. c1440. Prob. 94, p. 81 with plate on p. 82. (5, 4; 5). I have a colour slide of this. AR. c1450. Prob. 212, p. 98. (5, 3; 10) correctly solved. (Unusually, Vogel's notes, pp.160-161 & 211-213, say nothing about this problem.) Benedetto da Firenze. c1465. c1480. P. 106. Part of the text is lacking, but it must be (3,5;5). Calandri. Arimethrica. 1491. F. 63v. (2, 3; 5). Tagliente. Libro de Abaco. (1515). 1541. Prob. 130, ff. 60v61r. Philippo and Jacomo share lunch with Constanzo (50, 31; 40). Ghaligai. Practica D'Arithmetica. 1521. Prob. 27, ff. 66r66v. Three men have 3, 2, 1 loaves of bread and other foods worth 8, 6, 4. A fourth comes and shares with them, paying 9. How much should each of the three get? The total value of the food is 4x9=36, so the bread is worth 36 8 6 4 = 18, or 3 per loaf. So the first should get 9, the second 3 and the third owes them 2 !! Cardan. Practica Arithmetice. 1539. Chap. 66, section 16, ff. CC.iv.v CC.v.v (pp. 140141). Three men with bread, wine and fish in the amounts: 3,4,0;0,5,6;2,0,7; which are considered of equal value. A fourth man with one bread comes and they share the meal and the fourth man pays 5. Tartaglia. General Trattato. 1556. Book 12, art. 33, pp. 199r199v. Three men have quails and bread. The first has 6 quails and 2s worth of bread; second has 4 quails and 3s worth of bread; third has 2 quails and 5s worth of bread. They share with a fourth person who pays 8s. Buteo. Logistica. 1559.  x  thProb. 3, pp. 201202. Four share their food. First has 4 breads and 20 carrots; second has 1 bread and 32p of wine; third has 7 breads and 8 carrots; fourth has a cheese. If all have equal value, what is the value of each item? Prob. 4, pp. 202203. Three share their food with a fourth. First has 2 breads and 7 nummos worth of fish; second has 4 breads and 5 nummos worth of condiments; third has 1 bread and 8 nummmos worth of wine. Fourth pays 12 nummos for his share.  x  tD. Adams. Scholar's Arithmetic. 1801. P. 210, no. 6. (3, 5; 8). Doesn't give any attempts at division, nor a solution. Jackson. Rational Amusement. 1821. Arithmetical Puzzles, no. 12, pp. 34 & 5354. (3,5;8). Says it appears in an Arabian manuscript. The man with five loaves divides 5, 3; the other protests and divides 4, 4; judge divides 7, 1. Why? = Magician's Own Book (UK version), 1871, Arithmetical paradox, pp. 2829. John Badcock. Domestic Amusements, or Philosophical Recreations. Op. cit. in 6.BH. [1823]. Pp. 186187, no. 255: Arithmetical paradox. Same as Jackson, saying it appears in an Arabic manuscript. The New Sphinx. c1840. (3, 5; 8). Three Greeks. Extra person divides his money 3, 5, but the second was dissatisfied and had the matter referred to Solon, who gave the right division. Magician's Own Book. 1857. The three travellers, pp. 225226. (3, 5; 8). = Boy's Own Conjuring Book, 1860, pp. 195-196. Vinot. 1860. Art. LX: Chacun son )cot, p. 77. (3, 5; 4). Edward Brooks. The Normal Mental Arithmetic A Thorough and Complete Course by Analysis and Induction. Sower, Potts & Co., Philadelphia, (1858), revised, 1863. Further revised as: The New Normal Mental Arithmetic: A Thorough and Complete Course by Analysis and Induction; Sower, Potts & Co., Philadelphia, 1873. Several ordinary examples and some unusual examples.  x  th1863 p. 146, no. 17; 1873 p. 154, no. 15. (6, 10; 16) but the third person eats 4 more than each of the other two eat. 1863 p. 146, no. 19; 1873 p. 155, no. 17. (5, 9; 24) but C eats twice as much as B, who eats twice as much as A.  x  tBachetLabosne. Problemes. 3rd ed., 1874. Supp. prob. I: Deux Arabes allaient d3ner: ..., 1884: 181. (3, 5; 8). Mittenzwey. 1880. Prob. 61, pp. 1213 & 64; 1895?: 67, pp. 17 & 66; 1917: 67, pp. 16 & 63. (6, 4; 10). One suggest dividing 6, 4, the other suggests 5, 5. Gives correct solution. Hoffmann. 1893. Chap IV, no. 74: The three Arabs, pp. 165 & 220 = HoffmannHordern, p.147. (3, 5; 8). First says to divide 3, 5; second says 4, 4; third says both are wrong. Mr. X [cf 4.A.1]. His Pages. The Royal Magazine 10:6 (Oct 1903) 530531. The stranger's dinner. (3,5; 8). Refers to Arabians. Similar to Jackson. Pearson. 1907. Part II, no. 138, pp. 141 & 218. (3, 5; 8). Ball-FitzPatrick. Footnote cited in 7.G.1. 1908. Says the problem is Arabic, but gives no reason. Clark. Mental Nuts. 1916, no. 22. Real estate. A invests $5000, B invests $3000. They buy three houses of equal value. They each take one and then sell the third for $8000. How do they divide the money? Answer is $7000 and $1000. His 1904, no. 25; 1916, no. 39 is a standard version of (3, 5; 8) with sandwiches. Hummerston. Fun, Mirth & Mystery. 1924. A partnership problem, Puzzle no. 3, pp. 19 & 172. (10, 14; 6), then the second uses his receipts to buy more which are shared equally and his colleagues pay him how many more could he now buy? Kraitchik. La Math)matiques des Jeux. Op. cit. in 4.A.2. 1930. Chap. 1, pp. 7-8: Probl/me des Pandects. He gives several examples and says they come from Unterrichtsbltter fGr Mathematik und Naturwissenschaften 11, pp. 81-85, ??NYS.  x  thNo. 22: (3, 5; 8). No. 23: (2, 3; 3). No. 24: (11, 14, 17; 42). No. 25: (5, 7; 12). No. 26: (7, 8; 30). Caius and Sempronius share 7 and 8 with Titus who paid them 14 and 16. Sempronius protests and a judge divides it as 12 and 18. "C'est probablement cette version qui a donn) ! ce probl/me le nom de celui de Pandectes." No. 27: (3, 5; 80). No. 28: (3, 4; 7) cultivating fields. No. 29: (90, 120; 70) digging a ditch. No. 30: (3, 4; 4). No. 31: (2, 3, 6, 9; 4) heating a workshop.  x  tRohrbough. Brain Resters and Testers. c1935. The Travelers' Dinner, pp. 2324. Arabs, (3,5,; 8). McKay. Party Night. 1940. No. 25, p. 182. A & B give a party and invite 2 and 3 guests. The party costs 35s how do they divide the expense? Initial reaction is in the ratio 2 : 3, but it should be 3 : 4. (Also entered in 7.H.5) Kraitchik. Mathematical Recreations. Op. cit. in 4.A.2. 1943. The problem of the Pandects, pp. 28-29. c= No. 26 of Math. des Jeux. Sullivan. Unusual. 1943. Prob. 19: An Arab picnic. (3, 5; 8). 7.H.4.EACH DOUBLES OTHERS' MONEY TO MAKE ALL EQUAL, ETC. See 7.R for related problems. See Tropfke 647-648. Diophantos. Arithmetica. c250. Book I.  x  thNo. 22, p. 138. "To find three numbers such that, if each give to the next following a given fraction of itself, in order, the results after each has given and taken may be equal." Illustrates with fractions 1/3, 1/4, 1/5. No. 23, pp. 138-139, is the same with four numbers, illustrated with fractions 1/3,1/4,1/5, 1/6.  x  tMahavira. 850. Chap. VI, v. 259-267, pp. 160-162.  x  th260: 3 men, each doubles the next to make all equal. 262: same with 5 men. 263: each 3/2's the next. Also each 5/3's the next. 266: each doubles others. Also each 3/2's others.  x  tFibonacci. 1202. Pp. 287-293 (S: 409415) gives several versions.  x  thPp. 287-288 (S: 409410). w 5/2's x; x 7/3's y; y 9/4's z; z 11/5's x and then all are equal. Answer: 5862, 2858, 2760, 2380 and he notes that these can be multiplied by any value. Pp. 288-291 (S: 410413). Same operations, but the results are in the proportion 5:4:3 : 2. Answer: 22875, 10000, 8355, 7280, and he also divides these by 5. Pp. 291-292 (S: 413414). w 5/2's all the others, etc. with the above ratios, then all are equal. Answer: 8436, 3288, 1440, 696. Pp. 292-293 (S: 414415). Same as the previous, but with results in proportion 5:4:3: 2. Answer: 29706, 11568, 498, 2256. (He erroneously has 29826 for the first value.)  x  tFibonacci. Flos. c1225. In Picutti, pp. 326332, numbers VIIIX.  x  thPp. 243244: De quatuor hominibus qui invenerunt bizantios. First man doubles the second's money, then the second man triples the third's, ..., to make all equal. Answer: 89, 77, 47, 27 with total of 240. Pp. 244245: Above continued. First doubles the others, second triples the others, ..., to make all equal. Answer: 241, 161, 61, 17 with total 480. Pp. 246247: Questio similis suprascripte de tribus hominibus. First 5/2's the others, second 10/3's the others, third 17/4's the others to make all equal. Answer: 1554, 738, 258 with total 2550 and he then divides through by 6.  x  tLucca 1754. c1330. Ff. 61v-62r, p. 142. 4 & 3 men. Each doubles the others to make all equal. In the 4 man case, he specifies the total money is 400. Giovanni di Bartolo. Op. cit. in 7.H. c1400. Prob. 9, pp. 17-18. Four men, each doubles the others' money and the product of the results is 1000. He assumes, for no clear reason, that the original amounts are proportional to 8 : 4 : 2 : 1. AR. c1450. Prob. 231, pp. 107-108 & 169-171. Two men, each doubles the others' money and then both have 13. Muscarello. 1478. Ff. 78r78v, pp. 193194. Four men, each doubles the others' money, then all are equal. Answer: 33, 17, 9, 5. Chuquet. 1484. Prob. 148. 3 people. Mentioned in passing on FHM 230. Calandri. Aritmetica. c1485. Ff. 101r102r, pp. 202-204. 6 men. Each doubles the others to make all equal. Pacioli. Summa. 1494.  x  thF. 105r, prob. 16. First wins 1/2 of the second's; second wins 1/3 of the third's; third wins 1/4 of the first's to make all equal 100. I get (200, 400, 300)/3. See Tonstall for corrections and intended interpretation. Pacioli gives no working and just states answers that are printed differently in the two editions, but from Tonstall we see that they are intended to be: 55 5/9 (given as 44 4/9 and as 1444/9), 1111/9, 133 1/3. F. 189r, prob. 6. First gives 7/12 of his money to the second, who then gives 11/30 of his money to the first, when both are equal. He gives no working and just one answer: 70/9, 658/95. I find the general answer is 47x = 48y.  x  tTonstall. De Arte Supputandi. 1522. P. 245. First wins 1/2 of the second's; second wins 1/3 of the third's; third wins 1/5 of the first's to make all equal 100. This is a correct version of Pacioli. He gives (500, 1000, 1200)/9 and shows the calculation which implies all three calculations are done at once that is, the 1/5 of the first's money is based on what he had to start, not what he has after winning from the second. Cardan. Practica Arithmetice. 1539. Chap. 66, section 91, ff. GG.viii.v HH.i.v (pp.165-166). A situation somewhat similar to 7.P.7, where the first two take money leaving the third with 5. Friend says the first is to give 10 and @ of what he has left to the second and the second is then to give 7 and  of what he has left to the third to make their amounts proportional to (3, 2, 1). Answer is x = 172, y = 39 and the total sum is 216. Tartaglia. General Trattato, 1556, art. 11, 18, 37, 38, pp. 240v, 241v, 244v-245r. 2 and 3 people versions. Buteo. Logistica. 1559. Prob. 63, pp. 270273. Three players first wins 1/2 of second's, second wins 1/3 of third's, third wins 1/4 of what the first had originally, and all wind up with 100. That is, we have x+y/2-x/4=y-y/2+z/3=z-z/3+x/4=100. Bachet. Problemes. 1612. Addl. prob. VIII, 1612: 154160; 1624: 226-233; 1884: 162-167. 3people; also with tripling. Labosne adds the general case. van Etten. 1624. Prob. 57 (52), pp. 52-53 (78). 3 people version used as a kind of divination. Ozanam. 1694. Prob. 26, 1696: 81; 1708: 72. Prob. 47, 1725: 253. Prob. 17, 1778: 209; 1803: 204. Prob. 16, 1814: 177; 1840: 91. Three person version, resulting in each having 8, used as a kind of divination. Euler. Algebra. 1770. I.IV,IV.616: Question 4, pp. 211-212. Three players, all winding up with 24. Hutton. A Course of Mathematics. 1798? Prob. 42, 1833: 223; 1857: 227. Five players, all ending up with 32. Silvestre Fran'ois Lacroix. (l)mens d'Alg/bre, a l'Usage de l'(cole Centrale des QuatreNations. 14th ed., Bachelier, Paris, 1825. Section 82, ex. 9, p. 123. Three players, each doubles others, all ending with 120. Thomas Grainger Hall. The Elements of Algebra: Chiefly Intended for Schools, and the Junior Classes in Colleges. Second Edition: Altered and Enlarged. John W. Parker, London, 1846. Pp. 130131, ex. 16. Each 3/2's others, with total given as 162. Cf Mahavira 266. G. Ainsworth & J. Yeats. A Treatise on the Elements of Algebra. H. Ingram, London, 1854. Exercise XXXVIII, pp. 81-83 & 178.  x  thNo. 14: A doubles B, B doubles A, A doubles B, to make both have 80. How much originally? Answer is 100, 50 instead of 110, 50. No. 15: same as no. 14, with one more stage to make both equal n. No. 28: usual problem with four gamblers, ending with 64. No. 29: usual problem with 7 baskets of apples, all ending with 128. No. 30: usual problem with n persons, all ending with a. Solution is badly misprinted the ith should start with an amount a(2ni + 1)/n 2n.  x  tVinot. 1860. Art. LXXI: Les trois joueurs, pp. 8687. Three person version, all ending up with 24. Boy's [Own] Magazine 2:5 (No. 11) (Nov 1863) 459 [answer would be in Jan 1864, ??NYS]. (Reprinted as (Beeton's) Boy's Own Magazine 3:11 (Nov 1889) 479. Mathematical question 137. Three boys playing. Each pays the winner half of what he has. Each one wins once and then they have 30d, 60d, 120d. How much did they have to start? [In fact, they had the same amounts. In general, if they start with 1, 2, 4, then the amounts after the first, second, third wins are: 4, 1, 2; 2, 4, 1; 1, 2, 4.] Todhunter. Algebra, 5th ed. 1870. Examples XIII, no. 26, pp. 104 & 578. 3 men, each doubles the others to make all equal to 16. Mittenzwey. 1880.  x  thProb. 113, pp. 23 & 76; 1895?: 131, pp. 27 & 78; 1917: 131, pp. 25 & 75. =Ainsworth & Yeats, no. 14, except he doesn't say what the final result is. Solution is 110, 50. Prob. 126, pp. 26 & 76; 1895?: 144, pp. 30 & 79; 1917: 144, pp. 27 & 77. Four gamblers, each doubles the others, winding up with 64. 1895?: Prob. 76, pp. 18 & 68; 1917: 76, pp. 17 & 64. Three piles, each is used to double the next, making all have 8. 1895? just states the answer; 1917 sets up and solves the equations.  x  tLucas. L'Arithm)tique Amusante. 1895. Prob. XLIII: Les jouers, pp. 184185. Three players winding up with 12 each. Does general solution for n players winding up with a each. Workman. Op. cit. in 7.H.1. 1902. Section IX (= Chap. XXXI in c1928 ed.), examples CXLV, prob. 45, pp. 428 & 544 (434 & 577 in c1928 ed.). Version with five people. Hermann Schubert. Beispiel-Sammlung zur Arithmetik und Algebra. 3rd ed., G?schen, Berlin, 1913. Section 17, no. 107, pp. 66 & 140. Three people. Each gives away half his money to be equally shared by the others, and then they all have 8. Solution: 4,7,13. Loyd. Cyclopedia. 1914. Sam Loyd's Mystery Puzzle, pp. 226 & 369. (=MPSL2, prob. 85, pp. 61 & 148. = SLAHP: An initiation fee, pp.63 & 109.) 3 people version, resulting in the first person having lost 100. Collins. Book of Puzzles. 1927. The five gamblers puzzle, p. 75. They all end up with $32. 7.H.5.SHARING COST OF STAIRS, ETC.  tSee Tropfke 529. The simplest form is given by Sridhara, BR, Pseudodell'Abbaco, Gori, von Schinnern.  t Mahavira. 850. Chap. VI, v. 226-232, pp. 151-153.  x  th227. Porter carrying 32 jack-fruits over distance 1 will receive 7 of the fruits. He breaks down at distance . How much is he due? Rule says x/32*1=(7-x)/(32 - x)(1 - ), i.e. the wages per fruit-mile should be the same for both parts of the journey. (Properly, this problem leads to an exponential.) 229. Porter carrying 24 jack-fruits for distance 5 will earn 9 of them. Two porters share the work, the first earns 6 and the second earns 3. How far did the first one carry the fruits? 232. Twenty men are to carry a palanquin a distance of 2 for wages of 720. But two men drop off after distance ; three more drop off after another  and five more drop off after half the remaining distance. How much does each earn? He says each quarter of the distance is worth 180 and divides this equally among the carriers for each quarter.  x  tSridhara. c900.  x  thV. 67(ii), ex. 84-85, pp. 53 & 95. Porter carrying 200 palas of oil for 5 panas wages. But the bottle leaks and only 20 palas remain at the end. How much should he be paid? Rule says to pay (20 + 180/2)/200 of the wages. V. 68, ex. 86-90, pp. 53-55 & 95. Ex. 86-87. Four men watch a dance for , ,  and all of a day. The dancers' fee is 96. How much should each pay? He charges 24/4 per watcher for the first quarter, 24/3 per watcher for the second quarter, ..., giving payments 6,14,26,50. Ex. 88. Ten men are to carry a palanquin a distance of 3 for wages 100. Two men drop off after distance 1 and another three after a total distance of 2. How much does each earn? Divides as in Mahavira's 232. Ex. 89. Five chanters perform 1, 2, 3, 4, 5 chants for a fee of 300. How much does each earn? Each chant earns 60, divided among the chanters of it. V. 70-71, ex. 92-94, pp. 56-58 & 96. Ex. 92. Porter carrying 24 jack-fruits for distance 5 will earn 9. What does he earn for carrying distance 2? Ex. 93-94. Porter carrying 24 jack-fruits for distance 5 will earn 9 of them. Two porters split the carrying, the first earns 4 and the second earns 5. How far did they carry? Solutions of these are based on the rule in Mahavira 227.  x  tBR. c1305. No. 36, pp. 54-57. Divide oil among 12 lamps which are to burn 1, 2, ..., 12 hours. Pseudodell'Abbaco. c1440. Prob. 92, pp. 78-81. House rented to 1 person the first month, who shares with a 2nd the 2nd month, who share with a 3rd the 3rd month, ..., who share with a 12th the 12th month. How much does each pay? He says many obtain 12/78,11/78, ..., 1/78, but that it is more correct if the first pays (1+1/2+1/3+...+1/12) * 1/12, the second pays (1/2+1/3+... + 1/12) * 1/12, ..., the 12th pays 1/12 * 1/12. Gori. Libro di arimetricha. 1571. F. 73r (p. 80). Same as Prob. 92 of Pseudodell'Abbaco, but with only the first solution. Bullen. Op. cit. in 7.G.1. 1789. Chap. 38, prob. 32, p. 243. Four men hire a coach to go 130 miles (misprinted as 100). After 40 miles, two more men join. How much does each pay? Clemens Rudolph Ritter von Schinnern. Ein Dutzend mathematischer Betrachtungen. Geistinger, Vienna, 1826, pp. 14-16. Discusses general problem of sharing a cost of n for lighting a x floor staircase. Does the case n = 48, x = 4, getting 3, 7, 13, 25, which are the same proportions as Sridhara, ex. 86-87. Dana P. Colburn. Arithmetic and Its Applications. H, Cowperthwait & Co., Philadelphia, 1856. Miscellaneous Examples No. 66, p. 365. A and B hire a horse and carriage for $7 to go 42 miles and return. After 12 miles, C joins them, and after 24 miles, D joins them. How much should each pay? No solution is given, but there is a Note after the question saying there are two common ways to do the allocation. First is in proportion to the miles travelled, so A : B : C : D = 42 : 42 : 30 : 18. Second is to divide the cost of each section among the number of riders, which here gives A:B:C:D=29:29:17 : 9 Clark. Mental Nuts. 1897, no. 95; 1904, no. 71; 1916, no. 14. The livery team. I hire a livery team for $4 to go to the next city, 12 miles away. At the crossroads 6 miles away, I pick up a rider to the city who then rides back to the crossroads. How much should he pay? Answer is $1. M. Adams. Puzzle Book. 1939. Prob. C.135: Answer quickly!, pp. 158 & 187. Man hires car to go to theatre. He picks up and drops off a friend who lives half way to the theatre. How do they divide the fare? Answer is 3: 1. Depew. Cokesbury Game Book. 1939. Passenger, p. 219. Similar to M. Adams. McKay. At Home Tonight. 1940. Prob. 13: Sharing the cost, pp. 65 & 79. Similar to M.Adams. McKay. Party Night. 1940. No. 25, p. 182. A & B give a party and invite 2 and 3 guests. The party costs 35s how do they divided the expense? Initial reaction is in the ratio 2 : 3, but it should be 3 : 4. (Also entered in 7.H.3.) Doubleday 3. 1972. Fair's fair, pp. 6162. Man hires a taxi to go to the city and pays in advance. Halfway there, he picks up a friend. Later they go back, with the friend dropped at the halfway point. How should they share the fare? Answer says the friend should pay one third, because the man 'had only hired the taxi to take him into town ... not for a roundjourney.' This may be introducing the following extra feature. Normally the friend would pay 1/4 of the total cost. But when the taxi has got halfway back, the fare shown will only be 3/4 of the total cost and the friend should pay 1/3 of that amount. The problem does say that the same driver has been used and there was no charge for the waiting time. But I wonder whether taximeters can be stopped and restarted in this way. It would be more natural if they caught another taxi back. Then at the halfway point, the fare shown would be 1/4 of the total cost and the friend show pay all of the amount shown!  7.H.6.SHARING A GRINDSTONE  tNew section. I have seen other examples. A grindstone of radius R is to be shared between two (or k) buyers one grinding until his share is used. An inner circle of radius r is unusable. The first man should grind to radius x where x2 r2 = (R2 r2)/k or x2Ġ=[(k1)R2 + r2]/k. It is straightforward to adapt this to the case when the buyers contribute unequal amounts to the purchase price. Clark gives a problem of sawing through a tree which uses the fact that the area of a segment of a circle of radius R and segment height H is R2cos-1(R-H)/R-(R-H)(2RH-H2). Though wellknown, this seems about on the border of what I consider to be recreational.  t Anonymous proposer; solution lacking. Ladies' Diary, 170910 = T. Leybourn, I: 56, quest. 9. [??NX of p. 6.] Share a grindstone among seven people. 2R = 60". Carlile. Collection. 1793. Prob. XXIV, p. 16. Three men buy a grindstone of radius 20 for 20s. They pay 9s, 6s, 5s respectively. How much should each man get to grind? He makes no allowance for wastage. HuttonRutherford. A Course of Mathematics. 1841? Prob. 22, 1857: 558. Share a grindstone among seven people. 2R = 60". He takes r = 0. Dana P. Colburn. Arithmetic and Its Applications. H, Cowperthwait & Co., Philadelphia, 1856. Miscellaneous Examples No. 43, p. 362. Four men sharing a grindstone 4 ft in diameter. No indication of inner wastage. No solution given. Clark. Mental Nuts. 1904, no. 93. Cutting trees. Three men cutting through a tree of diameter three. One cuts in one from one side, the next cuts in one from the opposite side. How much is left? Answer is 41.64%, which is correct. [The obvious question is how far should the first two men get so that all three cut an equal area? I find it should be .7350679153972 of the radius.] Collins. Fun with Figures. 1928. A grindstone dispute, p. 22. 2R=5ft 6 in; 2r = 18 in. 7.H.7.DIGGING PART OF A WELL.  tRecently separated from 7.H.5. See Tropfke 529. The 'digging a well' problem has a contract to dig a well a deep for payment b, but the digging stops at c. How much should be paid? The value of b is not always given and then only the ratio of the part payment to the total payment is sought. NOTATION this problem is denoted (a, b; c). If the difficulty is proportional to depth, then integration yields that the payment should be proportional to (a/c)2. A common medieval approach is use the proportion 1+...+c:1+...+ a. We let TaĠ = 1 + ... + a, so this proportion is Tc : Ta. Benedetto da Firenze; della Francesca; Calandri, 1491, are the only cases where c>a. Della Francesca begins the inverse problem if the contract for depth a is worth b and work stops at x such that the value of the dug hole is d, what is x? Denote this situation as (a,b;x) worth d. See: della Francesca; Pacioli; Calandri, Raccolta; Buteo. Ozanam, Vyse and Jackson are the only ones to consider use of other arithmetic progressions. Berloquin gives a simple argument that work is proportional to a2/2.  t In Neugebauer & Sachs, op. cit. in 7.E, the problems discussed on pp. 8191 involve digging out ditches and the cost or difficulty of digging increases with the depth, but none of these are like the problem considered here, though Tropfke 529 notes a resemblance. Tabari. Mift]h almu]mal]t. c1075. ??NYS quoted and discussed by Tropfke 529.  x  thP. 96, part III, No. 17. Calculation of ditchwork. (15, 30; 10). Tropfke doesn't give a solution but says it is similar to the following. P. 227, part VI, no. 61. "Apportionment of wells and cisterns. In order to dig a well, the earth must be lifted out. For the first ell, the earth comes up one ell; for the second ell, the earth comes up two ells; for the third, three ells; etc. until the end. For this calculation, one must use the series of of natural numbers. So we take 1 as first term, to which 2 as second term gives 3, to which 3 as third term gives 6 as sum, etc. In this way, we do each time, and that is the 'bast' (literally 'extension'). Example: the well depth is 10 ells; what is the 'bast'? 1 + 2 = 3, 3 + 3 = 6, ..., 45+ 10 = 55. That is the 'bast' of a well 10 ells deep." So he divides in the standard ratio T10/T15Ġ=55/120=1/3+1/8. "We multiply that by 30; this yields 13 [+] 1/2 [+] 1/4. That is the payment for a well of 10 ells, when the other well costs 30 dirhems." Tropfke says the problem is reminiscent of a Babylonian one cf above.  x  tQazwini = al-Qazw3n3. Loc. cit. in 7.H.3. c1260. P. 253. German translation only man contracts to dig a well 10 ells deep for 10 dirhems. He stops at 9 ells, so we have (10, 10; 9). Man asks for 9 dirhems, but an expert says only 8 and somewhat more. BR. c1305. No. 22, pp. 40-43. Man contracts to dig 10 x 10 x 10 cistern but only does 5x5 x 5. Text gives him 1/8 of the value. Lucca 1754. c1330. F. 64v, p. 152. Man digging a well, (10, b; 8). He divides in ratio T8Ġ:T10 = 36 : 55. Pseudodell'Abbaco. c1440. Prob. 102, p. 87 with plate on p. 88. Man contracts to dig a well 20 deep and stops at 14, i.e. (20, b; 14). Author divides in ratio T14 : T20 = 1 : 2. But he says he doesn't think this is a correct method, though he doesn't know a better one. I have a colour slide of this. Benedetto da Firenze. c1465. Pp. 115-116. If a well 12 deep is worth 12, how much is a well 14 deep worth? This is (12, 12; 14). He takes values proportional to Td. Muscarello. 1478. Ff. 66v67r, pp. 176177. Man to dig a hole but hits water and has to stop, (10,b;7). Divides in the ratio T7Ġ:T10 = 28 : 55. della Francesca. Trattato. c1480. F. 52r (121122). Men agree to dig a well of depth 4 for 10, but no water is found and they continue until the cost is 11 more. I.e. (4, 10; x) worth 21. Since T4 = 10 and T6 = 21, they dig to 6. English in Jayawardene. Calandri. Arimethrica. 1491. F. 65v. (12, 12; 16). Pacioli. Summa. 1494.  x  thF. 40v, prob. 8. Dig a well, (11, 11; 6). Divides as T6Ġ:T11Ġ=7: 22, so the partial well is worth 7/2. F. 40v, prob. 9. Dig a well, (11, 11; x) worth 7/2. Part II, f. 55v, prob. 38. Dig a well, (10, 10; 6). Divides as T6Ġ:T10 = 21 : 55. Part II, f. 55v, prob. 39. Dig a well, (10, 10; x) worth 4. He notes 4 : 10 = 22 : 55, so we want n(n+1)/2 = 22, which would give n = (1 + 177)/2 = 6.15207.... He interpolates as 6 days plus 1/7 of the sixth day, i.e. n = 6.14286....  x  tCalandri. Raccolta. c1495. Prob. 38, pp. 33-34. If a well 24 deep is worth 24, how deep a well is worth 40? I.e. (24, 24; x) worth 40. He takes values proportional to Td. Tagliente. Libro de Abaco. (1515). 1541. Prob. 110, ff. 55r55v. Dig a well, (9, 24; 5). Divides in ratio T5Ġ:T9 = 15 : 45. Apianus. Kauffmanss Rechnung. 1527. F. D.vi.v. Mason to build a tower 100 high for 100. He falls ill after 84. Divides in ratio T84 : T100. Cardan. Practica Arithmetice. 1539. Chap. 66.  x  thSection 8, ff. CC.i.r CC.i.v (p. 138). Men dig a well, (34, 60; 20). Divides in ratio T20Ġ:T34 = 6 : 17. Section 10, ff. CC.i.v - CC.ii.v (p. 138). Man building a wall 5 high at prices 10,20,40, 80, 160 per unit of height. He stops at height 2. Takes half a unit of height as having 2 times the value of the previous halfunit, so the interval from 2 to 3 high worth 40 divides into two halves worth x and x2, giving x=40/(1+2).  x  tButeo. Logistica. 1559.  x  thProb. 35, pp. 238240. Dig a well, (100, 50; 50). Divides as T50Ġ:T100Ġ=1275:5050. Also does (200, 50; 100). His depths are initially in cubits, but are converted to 'semipedes' 120 cubits is 50 semipedes. Prob. 36, pp. 240241. Dig a well, (100, 50; x) worth 28 22/101.  x  tOzanam. 1694, 1725.  x  thProb. 7, question 1, 1696: 30; 1708: 27. Prob. 10, question 1, 1725: 6061. Prob.2,1778: 65; 1803: 6768; 1814: 60; 1840: 3233. Man digging a well 20 feet deep, to receive 3, 5, 7, ..., for each successive foot. [This is not really in this section, but is included because later ed. use it as the basis of the next problem.] Prob. 51, question 1, 1725: 256-257. Prob. 3, 1778: 6667; 1803: 6869; 1814: 6162; 1840: 33. Man digging a well, (20, b; 12) (1778 et seq. change 12 to 8). 1725 divides as T12Ġ:T20Ġ=78: 210. 1778 notes that the difficulty of the work increases in arithmetic progression, but that there are many such progressions. He then posits that the first unit is worth 1/4 when the agreed payment is 20 and this gives a difference of 30/11 for the arithmetic progression. If the cost per unit depth is an arithmetic progression: A,A+D,...,A+(a-1)D and  = D/A and d is the value of the partial well, then d/b = ((2AD)c + Dc2)/((2AD)a + Da2).  x  tLes Amusemens. 1749. Prob. 44, p. 176. Mason to dig a well, (10, b; 4). Divides as T4Ġ:T10 = 10:55. Vyse. Tutor's Guide. 1771? Prob. 17, 1793: p. 136; 1799: pp. 144145 & Key p. 188. Same as Ozanam, prob. 7, but with depth 30. Jackson. Rational Amusement. 1821. Curious Arithmetical Questions, no. 25, pp. 20 & 79. Dig 20 yards for 20. Man falls sick after 8 yards. "How much was then due to him, on a supposition that the labour increases in arithmetical proportion as the depth?" I.e. (20, 20; 8). Solution notes that the data does not determine what the arithmetic progression is and chooses 5s as cost of the first yard see Ozanam. Unger. Arithmetische Unterhaltungen. 1838. Pp. 182 & 263, no. 693. Man digging a well 49 feet deep. First foot costs 15, but each successive foot costs 6 more than the previous. Find cost of last foot and total cost. So this is really an arithmetic progression problem, but I haven't seen others of those using this context. Vinot. 1860. Art. CVII: Probl/me du puits et du Ma'on, pp. 126127, (20, 400; 10). He assumes the cost of digging up a unit depth is 5 and that lifting the ith unit raises it from its centre of gravity, so is given by A, 3A, 5A, ..., 39A, where A has to be determined from the total cost. He finds A = 3/4 and d = 125. Pierre Berloquin. The Garden of the Sphinx, op. cit. in 5.N. 1981. Prob. 117: A bailout fee, pp. 66 & 166. Man contracts to bail out a 20 yard deep well for $400 and gives up at 10 yards. Answer says value is proportional to the depth d times the average distance lifted, i.e. d/2, hence value is proportional to d2/2 and this is the result that integration produces. 7.I. FOUR FOURS, ETC.  tExpress an integer using four 4s, etc. Cupidus Scientiae, 1881, seems to be the first to ask for solutions to a lot of the integers, rather than a few specific examples. The next examples of the general form are Cunningham & Wiggins (1905), Pearson (1907), Ball (1911), Ball (1912). Dawson (1916) is the first to ask for four R's, where R is indeterminate, e.g. 3=(R+R+R)/R. I have included examples where a set of numbers and operations is given and one has to obtain a given value. This overlaps a bit with 7.I.1, where the object is to find the maximum possible value, and with 7.AC.36, where one uses all nine or ten of the digits and I have included problems of inserting signs into 12...9 to make 100 in 7.AC.3.  tDilworth. Schoolmaster's Assistant. 1743. Part IV: Questions: A short Collection of pleasant and diverting Questions, p. 168.  x  thProblem 4: "Let 12 be set down in four Figures and let each Figure be the same." Problem 9. "Says Jack to his brother Harry, I can place four threes in such manner that they shall just make 34; can you do so too?"  x  tLes Amusemens. 1749. Pp. 5254. Several problems leading to: 7 7/7, 33 3/3, 55 5/5, 999/9, 77 77/77, 2222 2222/2222, 11 1/1, etc. See the entry in 7.AN. Vyse. Tutor's Guide. 1771? Prob. 1, 1793: p. 155; 1799: p. 165 & Key p. 206. "Four Figures of nine may be so placed and disposed of as to denote and read for 100, neither more nor less. Pray how is that to be done?" Pike. Arithmetic. 1788. P. 350, no. 16. "Said Harry to Edmund, I can place four 1's so that, when added, they shall make precisely 12; Can you do so too?" Jackson. Rational Amusement. 1821. Arithmetical Puzzles.  x  thNo. 2, pp. 1 & 51. "It is required to express 100 by four 9's." No. 4, pp. 2 & 51. Use three 2s to make , 1 and 2. No. 5, pp. 2 & 51. Express 12 by four equal figures. No. 17, pp. 5 & 56. Express 6 by four 5s. Answer is: 5.5 + 5/5. No. 33, pp. 8 & 59. Use four 2s to make 1/8, 1/2, 2 and 8. No. 40, pp. 10 & 62. Use three 3s to make 1/3, 1 and 3. No. 42, pp. 10 & 62. Use four 3s to make 1/243, 1/27, 1/3, 3, 27 and 243. No. 43, pp. 10 & 62. Use five 3s to make the same numbers as in no. 42. No. 44, pp. 10 & 63. Express 78 by six equal digits.  x  tEndless Amusement II. 1826? Prob. 23, p. 201. "Put down four nines, so that they will make one hundred." Child. Girl's Own Book. Arithmetical puzzles, no. 5. 1832: 170 & 179; 1833: 184 & 193; 1839: 164 & 173; 1842: 282 & 291; 1876: 231 & 244. "Place four nines together, so as to make exactly one hundred. In the same way, four may be made from three threes, three may be made from three twos, &c." The 1833 solution is printed rather oddly as 1999-9, while the 1839 and 1842 solution is 99 99 and the 1876 solution is 99 99. Nuts to Crack III (1834), no. 211. "Write down four nines so as to make a hundred." Family Friend 1 (1849) 150 & 178. Problems, arithmetical puzzles, &c. 4. "Put four fives in such a manner, that they shall make 6. D.F." Answer is 5 5/5 + .5. = The Sociable, 1858, Prob. 46: A dozen quibbles, part 12: pp. 300 & 318. = Book of 500 Puzzles, 1859, prob. 46: part 12, pp. 18 & 36. Boy's Own Book. To place four figures of 9 in such a manner as to represent 100. 1855:601. Magician's Own Book. 1857. Quaint questions, p. 253. [No. 4] "Place three sixes together, so as to make seven." [No. 6] "Place four fives so as to make six and a half." [Boy's Own Conjuring Book, 1860, pp. 224225, has the Quaint Questions, but omits these two questions!] Book of 500 Puzzles. 1859.  x  thProb. 46: A dozen quibbles: part 12, pp. 18 & 36. As in Family Friend. Quaint questions, p. 67. [Nos. 4 & 6] Identical to Magician's Own Book.  x  tCharades, Enigmas, and Riddles. 1860: prob. 30, pp. 60 & 64; 1862: prob. 31, pp. 136 & 142; 1865: prob. 575, pp. 108 & 155. "Write a Hundred with 4 nines." (1862 & 1865 have slightly different typography.) Illustrated Boy's Own Treasury. 1860.  x  thProb. 1, pp. 427 & 431. "Put down four nines, so that they shall make one hundred." Prob. 26, pp. 429 & 433. "Put four fives in such a manner, that they shall make 6." Prob. 38, pp. 430 & 434. "It is required to place four 2's in such a manner as to form four numbers in geometrical progression?" Uses four 2s to make each of 1/8,1/2, 2, 8.  x  tLeske. Illustriertes Spielbuch fGr Mdchen. 1864?  x  thProb. 56415, pp. 253 & 395. Write 100 with 4 nines. Write 1000 with no zeroes answer: 999 9/9. Prob. 56418, pp. 253 & 395. Write 100 with no zeroes, using 4, 6 or 8 figures. Answers: 99 3/3; 99 44/44; 99 999/999.  x  tMagician's Own Book (UK version). 1871. Paradoxes [no. 3], p. 37. "With four fives make 6? 5 5/5 .5." where the 5/5 is written as a 5 over a 5 with no fraction bar. Cf Jackson and Magician's Own Book. Hugh Rowley. More Puniana; or, Thoughts Wise and OtherWhy's. Chatto & Windus, London, 1875. P. 300. "Write down one hundred with four nines." Mittenzwey. 1880. Prob. 1, pp. 1 & 58; 1895?: 1, pp. 7 & 62; 1917: 1, pp. 7 & 56. Write 100 with six equal digits. "Cupidus Scientiae" (possibly the editor, Richard A. Proctor). Four fours, singular numerical relation. Knowledge 1 (30 Dec 1881) 184, item 151. A bit vague as to what operations are permitted, but wants four 4s to make various values. Says he has not been able to make 19. H. Snell. Singular property of number 4. Knowledge 1 (6 Jan 1882) 209, item 178. 19=4!-4 4/4. Editor says 4! is not reasonable for the problem as posed. Solutions from various contributors. Four fours. Knowledge 1 (13 Jan 1882) 229, item 184. Numerous solutions for 1 through 20, except 19. Solutions for 19 are: 4/.4 + 4/.4; 4! 4 4/4; 4/.4-4/4 ("manifestly erroneous"); (4 + 4 .4)/.4; (x+x-.x)/.x in general. Four 3s give same results as three 5s, except for 17. Albert Ellery Berg, ed. Op. cit. in 4.B.1. 1883. P. 373. "Place three sixes together so as to make seven." Lemon. 1890.  x  thVagaries, no. 217(b), pp. 33 & 105. Three 6s to make 7. Arithmetical, no. 752, pp. 92 & 124. = Sphinx, no. 600, pp. 81 & 118. "Place four nines so as to make one hundred."  x  tBerkeley & Rowland. Card Tricks and Puzzles. 1892. Card Puzzles No. II: A curious addition sum, p. 2. Mentions "writing down 100 with four nines" as 99 9/9. (Sam Loyd.) One hundred pounds for correct answer to a puzzle. TitBits (14 Oct 1893) 25. "Find How to Arrange the Figures  4  5  6  7  8  9  0  in an Arithmetical Sum which Adds up the Nearest to 82." "Mr. Loyd is confident that no one will find it out." Indeed, Loyd will be paid 100 only if no correct answer is received. (Sam Loyd.) Solution of Mr. Sam Loyd's one hundred pound puzzle. TitBits (18 Nov 1893) 111. 805 + 97 + 46 = 82. (There are points over the 5, 9, 7, 4, 6, but my printer may not print these clearly.) Here the midline dot () is used for a decimal point. Because of the number of correct solutions, ten extra names were drawn from them for additional 5 prizes. "It seems that not a single person in the whole of America has sent the correct answer when a prize was offered there, but here we have received a very large number actually correct." [See MRE for another solution.] Report on the 82 puzzle appeared in 25 Nov and letter from Loyd appeared about two weeks later photocopies on order. Hoffmann. 1893. Chap. IV, no. 18: Another way to make a hundred, pp. 148 & 193 =HoffmannHordern, p. 120. Use six 9s to make 100. Ball. MRE, 3rd ed., 1896. P. 13. "... a question which attracted some attention in London in October, 1893, ...." [See Loyd above.] He says that the problem is to make 82 with the seven digits 9, 8, 7, 6, 5, 4, 0 and gives one solution as 8069 + 74 + 5 (with points over the 9, 4, 5 there should also be points over the 6 and 7). H. D. Northrop. Popular Pastimes. 1901. No. 10: A dozen quibbles, no. 12, pp. 68 & 73. Express 6 by four 5s. Answer is: 5 5/5 . 5, which seems pretty poor to me. c=Jackson, no. 17. Mr. X [cf 4.A.1]. His Pages. The Royal Magazine 10:5 (Sep 1903) 426427. "How to arrange four 9's to make 100." Ball. MRE, 4th ed., 1905. P. 14. Repeats the material in the 3rd ed of 1896, again omitting two points, and adds further questions. Use the 10 digits to total 1 a solution is 35/70 + 148/296 or to total 100 a solution is 50 + 49 + 1/2 + 38/76. Use the 9 digits to make four numbers which total 100 a solution is 78 + 15 + 29 + 364. A. Cunningham & T. Wiggins. ?? Math. Quest. Educ. Times 7 (1905) 43-46. ??NYS cited in Dickson I 460, item 45d. Expressions using four 9s and four 4s. Pearson. 1907.  x  thPart I, no. 43: The nimble nines, pp. 125 & 187. Verse asking for three 9s to make 16 solution is 96/6 !! Part II: On all fours, p. 107. Four fours in general, with a few examples.  x  tWehman. New Book of 200 Puzzles. 1908. P. 26. = Magician's Own Book, no. 4. Ball. MRE, 5th ed., 1911. Pp. 1314. Briefly restates the material in the 4th ed. as "questions which have been propounded in recent years. ... To the making of such questions of this kind there is no limit, but their solution involves little or no mathematical skill." "Another traditional and easy recreation .... I have never seen this recreation in print, but it seems to be an old and wellknown question." Deals just with four 4s and says one can get up through 170. G. N. Watson has pointed out that one can get further by using factorials and subfactorials. (The subfactorial of n is n=n![1/0!-1/1!+1/2! 1/3! + ... 1/n!].) The topic is not in earlier editions. W. W. Rouse Ball. Four fours. Some arithmetical puzzles. MG 6 (No. 98) (May 1912) 289-290. "An arithmetical amusement, said to have been first propounded in 1881, ...." [This would seem to refer to Knowledge, above.] Studies various forms of the problem. Says it occurs in his MRE see above. MRE 6th ed., 1914, p. 14, cites this article. Ball. MRE, 6th ed., 1914. Pp. 1314. He now splits the material into three sections. Empirical Problems. Restates the material in the 5th ed. as "... numerous empirical problems, ..." and omits Loyd's problem. "To the making of such questions there is no limit, but their solution involves little or no mathematical skill." He then introduces the "Four Digits Problem". "I suggest the following problem as being more interesting." Using the digits 1, 2, ..., n, express the integers from 1 up using four different digits and the operations of sum, product, positive integral power and base10 notation (or also allowing iterated square roots and factorials). With n=4, he can get to 88 or to 264. With n = 5, he can get to 231 or 790. Using 0,1, 2, 3, he can get to 36 (or 40). Under Four Fours Problem, he discusses what operations are permitted and says he can get to 112, or to 877 if subfactorials are permitted (citing his MG article for this). Mentions four 9s and four 3s problems. Williams. Home Entertainments. 1914. The six 9's, p. 119. "Express the number 100 by means of six 9's." Thomas Rayner Dawson. 1916. ??NYS. Cited in: G&PJ 3 (Jan 1988) 45 & 4 (Mar 1988) 61. Asks for four R's, where R is indeterminate, e.g. 3 = (R+R+R)/R. Ball. MRE, 7th ed., 1917. Pp. 1314. The material of the first two sections is repeated, but under "Four Fours Problem", he discusses the operations in more detail. With +, , x, , brackets and base10 notation, he can get to 22. Allowing also finitely iterated square roots, he can get to 30. Allowing also factorials, he can get to 112. Allowing also integral indices expressible by 4s and infinitely iterated square roots, he can get to 156. Allowing also subfactorials, he can get to 877. (In the 11th ed., 1939, pp. 1516, two footnotes are added giving expressions for 22 in the first case and 99 in the third case.) Gives some results for four 2s, four 3s, four 5s, four 9s. Mentions the general problem of n ds. Smith. Number Stories. 1919. Pp. 112-113 & 140-141. Use four 9s to make 19, 2 and 20. Ball. MRE, 9th ed., 1920. Pp. 1314. In the "Four Digits Problem", he considers n = 4, i.e. using 1, 2, 3, 4, and discusses the operations in more detail. Using sum, product, positive integral power and base10 notation, he can get to 88. Allowing also finitely iterated square roots and factorials, he can get to 264. Allowing also negative integral indices, he can get to 276. Allowing also fractional indices, he can get to 312. He then mentions using 0, 1, 2, 3 or four of the five digits 1, ..., 5. Under "Four Fours Problem", he repeats the material of the 7th ed., but adds some extra results so he has results for four ds, d = 1, 2, 3, 5, 6, 7, 8, 9. Ball. MRE, 10th ed., 1922. Pp. 1314. In the "Four Digits Problem", he repeats the material of the 9th ed., but at the end he adds that using all of the five digits, 1,..., 5, he has gotten to 3832 or 4282, depending on whether negative and fractional indices are excluded or allowed. Hummerston. Fun, Mirth & Mystery. 1924. Some queer puzzles, Puzzle no. 76, part 2, pp.164 & 183. "Express 100 by using the same figure six times." Dudeney. MP. 1926. Prob. 58: The two fours, pp. 23-24 & 114. = 536, prob. 109, pp. 34 & 248-249, with extensive comments by Gardner. King. Best 100. 1927.  x  thNo. 44, pp. 2021 & 48. = Foulsham's no. 14, pp. 8 & 11. "Can you put down four fifteens so that they come to 16,665? No. 45, p. 21 & 49. "Arrange the figures 1 to 7 so that they will amount to 100, when added together." Arrange four 9s to make 100. Gives two answers for the first part.  x  tHeinrich Voggenreiter. Deutsches Spielbuch Sechster Teil: Heimspiele. Ludwig Voggenreiter, Potsdam, 1930. P. 112. Write 1000 with seven or five equal digits. Perelman. FFF. 1934. 1957: probs. 98, 100 & 102, pp. 137 & 143144; 1979: probs. 101, 103 & 105, pp. 166167 & 174175. =MCBF, probs. 101, 103 & 105, pp. 167 & 176-178.  x  th101: Two digits: "What is the smallest integer that can be written with two digits?" 1/1= 2/2 = .... [Though I think 0/1 might be counted.] 103: Five 9's: "Write 10 with five 9's. Do it in at least two ways." 105: Four ways: "Show four different ways of writing 100 with five identical digits."  x  tPerelman. MCBF. 1937. Any number via three twos. Prob. 202, pp.398399. "A witty algebraic brainteaser that amused the participants of a congress of physicists in Odessa." n=-log2Ġlog2 n2, where n means n-fold iterated square root. HaldemanJulius. 1937. No. 16: Adding fives, pp. 5 & 21. Use four 5s to make 6. Answer is: 5 + 5/5 + .5. M. Adams. Puzzle Book. 1939. Prob. B.83: Figure juggling, part 3, pp. 78 & 107. Use a digit 8 times to make 1000. Answer uses 8s. Evelyn August. The BlackOut Book. Op. cit. in 5.X.1. 1939. Number, please!, pp. 20 & 210.  x  thUse the same odd figure five times to make 14. Four 9s to make 100. Four 5s to make 6.  x  tDepew. Cokesbury Game Book. 1939.  x  thThree eights, p. 216. Use three 8s to make 7. Twentyfour, p. 227. Use a digit three times to make 24. Answers: 33 3, 22 + 2.  x  tMcKay. Party Night. 1940. No. 9, p. 177.  x  th(a) Use three 9s to make 10. Answer: 9 9/9. (b) Use four 9s to make 20. Answer: 9 99/9. (c) Use three 9s to make 100. Answer: 99.9 (or 99.9 for clarity). (d) Use two 9s to make 10. Answer: 9/.9 or 9.9 (or 9.9).  x  tMeyer. Big Fun Book. 1940. A half dozen equals 12, pp. 119 & 738. Use six 1s to make 12. Answer: 11 + 11/11. George S. Terry. The Dozen System. Longmans, Green & Co., NY, 1941. ??NYS quoted in: Underwood Dudley; Mathematical Cranks ; MAA, 1992, p. 25. What numbers can be expressed with four 4s duodecimally? About 5 dozen. How many numbers can be expressed using each of the digits 1, 2, 3, 4 once only, again duodecimally? About 9 dozen and nine. Terry (or Dudley) also gives the results for decimal working as 22 and 88. Sullivan. Unusual. 1943. Prob. 17: Five of a kind. Write 100 with the same figure five times "and the usual mathematical symbols". Says it can be done with 1s, 2s, 5s (two ways), 9s and perhaps others. J. A. Tierney, proposer; Manhattan High School of Aviation Trades, D. H. Browne, H.W.Eves, solvers. Problem E631 Two fours. AMM 51 (1944) 403 & 52 (1945) 219. Express 64 using two 4s. Vern Hoggatt & Leo Moser, proposers and solvers. Problem E861 A curious representation of integers. AMM 56 (1945) 262 & 57 (1946) 35. Represent any integer with p a's, for any p  3 and any ac 1. Solution for n uses log to base ...a, with n radicals. S. Krutman. Curiosa 138: The problem of the four n's. SM 13 (1947) 47. Sullivan. Unusual. 1947. Prob. 28: A problem in arithmetic. What is the smallest number of eights which make 1000? G. C. S[hephard, ed.] The problems drive. Eureka 11 (Jan 1949) 1011 & 30.  x  thNo. 4. Use 1,2, 3, once each to make 19. Answer: (2/.1) [3]. Ibid. 12 (Oct 1949) 17 gives a simpler answer: (1 + 3!!/2). No. 7/ Use four 1s to express 7, 37, 71, 99. Answers: (1+1+1)! + 1; 111 x .1 [.1 is .111..., but may not show up clearly]; .1 x ((1/.1)!! 1 [same comment on .1]; 1/(.1 x .1) 1.  x  tAnonymous. The problems drive. Eureka 13 (Oct 1950) 11 & 2021.  x  thNo. 4: Start with 2 and use cubing and integral part of square root to form any positive integer. m cubings, followed by n roots gives 2^(3m/2n) = 2^(2man), where a= log2 3. Since a is irrational, we can choose ma n so that 2^(2man) is arbitrarily close to N + , so the integer part of it is N. No. 6: Use four 4s to approximate !. They get 3.14159862196..., using a ninefold root.  x  tAnonymous. The problems drive. Eureka 17 (Oct 1954) 89 & 1617. No. 5. Use four 4s to express 37; 57; 77; 97; 123. D. G. King-Hele. Note 2509: The four 4's problem. MG 39 (No. 328) (May 1955) 135. n=log[{log 4}/{log n 4}]/log 4 expresses any positive integer n in terms of three 4s. A slight variation expresses n in terms of four x's, for any real x c 0, 1. 1can be used by taking x = .1. He also expresses n in terms of m x's for real x c 0, 1 with m > 5 and also with m = 5. Anonymous. Problems drive. Eureka 18 (Oct 1955) 1517 & 21. No. 6. Use four 4s to make 7; 17; 37; 3,628,800. Anonymous. Problems drive, 1958. Eureka 21 (Oct 1958) 1416 & 30. No. 10. Use 1, 2, 3, 4, 5, in order to form 100; 3 1/7; 32769. Philip E. Bath. Fun with Figures. Op. cit. in 5.C. 1959.  x  thNo. 12: One hundred every time, pp. 10 & 41. Make an arrangement of x's which gives the result 100 for x = 1, 2, ..., 9. Answer: xxx/x xx/x. No. 20: Form fours, pp. 12 & 42. Eight 4s to make 500. No. 74: Signs wanted, pp. 28 & 54. Insert signs (+, , x, /) into a row of four x's to make 10 x, for x = 2, ..., 9.  x  tM. R. Boothroyd & J. H. Conway. Problems drive, 1959. Eureka 22 (Oct 1959) 1517 & 2223. No. 3. Use three 1s to make the integers from one to twelve, using only arithmetic symbols. (No trigonometric functions or integer parts allowed.) Young World. c1960. P. 54. Use five 9s to make 1000. B. D. Josephson & J. M. Boardman. Problems drive 1961. Eureka 24 (Oct 1961) 2022 & 24. Prob. K. Use three 7s to express 1, ..., 11, using only arithmetic symbols. J. H. Conway & M. J. T. Guy. ! in four 4's. Eureka 25 (Oct 1962) 18-19. Cite Eureka 13 (1950). Note that !=[(-4/4)!]4, if non-integral factorials are allowed. Show that any real number can be arbitrarily well approximated using four 4s. R. L. Hutchings & J. D. Blake. Problems drive 1962. Eureka 25 (Oct 1962) 2021 & 3435. Prob. H. Use four identical digits to represent 100 in as many ways as possible, but not using representations which are independent of the digit used, like (5x5)/(.5x.5). The give eight examples, using 9, 5, 5, 4, 4, 3, 3, 9, 1, and say there are more. D. E. Knuth. Representing numbers using only one 4. MM 37 (1964) 308-310. Gardner. SA (Jan 1964) adapted as Magic Numbers, chap. 5. Cites Knowledge as the origin. Magic Numbers gives numerous other citations. Marjorie Bicknell & Verner E. Hoggatt. 64 ways to write 64 using four 4's. RMM 14 (Jan-Feb 1964) 13-15. Jerome S. Meyer. Arithmetricks. Scholastic Book Services, NY, 1965. Juggling numbers, no. 3, pp. 83 & 88.  x  thMake 100 with four 7s. 77/.77. Make 20 with two 3s. 3!/.3. Make 7 with four 2s. (2/.2)/2 + 2. Make 37 with six 6s. 6*6 + 66/66.  x  tRipley's Puzzles and Games. 1966. Pp. 1617, item 2. Use 13 3s to make 100. Steven Everett. Meanwhile back in the labyrinth. Manifold 10 (Autumn 1971) 14-16. (=Seven Years of Manifold 1968-1980; ed. by I. Stewart & J. Jaworski; Shiva Publishing, Cheshire, 1981, pp.64-65.) n = -4 * log4 log4 n4, where log4 means log to the base4 and n means n-fold iterated square root. This is a variant of King-Hele's form. This article is written in a casual style and seems to indicate that this formula was devised by Niels Bohr. He gets a form for e, but it uses infinitely many factorial signs!!!!... Editor's note on p. 2 (not in the collection) gives an improvement due to Michael Gerzon, but it is unclear what is intended. The note gives another method due to Professor Burgess using sec tané1 m = (m+1) and 1 = nn which expresses n by one 1. [Henry] Joseph and Lenore Scott. Master Mind Pencil Puzzles. 1973. Op. cit. in 5.R.4. Numbersnumbers, part 3, pp. 109110. Use 13 3s to make 100. The give 33 + 33 + 33 + (3/3)3 + 3x3 + 3x3. I found 33 + 33 + 33 + 33/3 3x3 3/3, which seems simpler. Ball. MRE, 12th ed., 1974. Pp. 1517. Under "Four fours problem", the material of the 9th ed. and the footnotes mentioned at 7th ed are repeated, but the bound for four 9s is increased. Bronnie Cunningham. Funny Business. An Amazing Collection of Odd and Curious Facts with Some Jokes and Puzzles Too. Puffin, 1978. Pp. 38 & 142. Arrange three 9s to make 20. Answer: (9 + 9)/.9. Putnam. Puzzle Fun. 1978.  x  thNos. 5457: Ten is the number, pp. 10 & 35. Express 10 using five 9s, in four different ways. Nos. 5859: 3 + 3 + 3 = 30, pp. 10 & 35. Express 30 using three 3s, in two different ways. No. 97: Eight to one thousand, pp. 13 & 37. Use a digit eight times to express 1000.  x  tP. Grammer, I. McFiggans, N. Blacknell, T. Joyce, J. Anstey & A. Devonald. Counting in fours. MiS 9:4 (Sep 1980) 2122. Uses four 4s to express 1, ..., 50. Says 51 100 will appear in next issue, but they didn't. J. Bellhouse. Four fours. MiS 14:1 (Jan 1985) 15. Says the promised table for 51 100 (see Sep 1980 above) had not appeared, so his students found their own. Anne Williamson. 1985. MiS 14:4 (Sep 1985) 7. Use the four digits 1, 9, 8, 5 to express integers 1 100. Unhappy with expressions for 24, 31, 65 which use !. Ken Lister. Letter. MiS 15:2 (Mar 1986) 47. Responding to Bellhouse (Jan 1985). Corrects and improves some values, but says 71 and 73 have not been done. Expresses a/b, for single digits a, b, by use of four 4s. Angie Aurora. Letter. MiS 15:3 (May 1986) 48. Improvements for Williamson's problem Sep 1985 above. Joyce Harris. Letter: Four fours. MiS 15:3 (May 1986) 48. Responding to Lister (Mar 1986), gives expressions for 71 and 73. Bob Wasyliw. Letter: Four 4's the ultimate solution. MiS 15:5 (1986) 39. Adapts Everett's 1971 method to include non-positive integers. Simon Gray & Colin Abell. Letters: Four fours again. MiS 16:2 (1987) 47. Gray notes that 4 = 4 * 4, so that 'four 4s' is the same as 'at most four 4s'. He gives !=4*sin-1(4/4) and more complex forms. Abell gives ! = - (-4/4) * log (-4/4) [The first minus sign is ambiguous??] and ! = - (4*4) * Tané1(4/4) [The minus sign is wrong.] Tim Sole. The Ticket to Heaven and Other Superior Puzzles. Penguin, 1988. Number play (ii) - (iv), pp. 15, 29 & 178-180.  x  th(ii). Using +, -, *, /, ., ! and brackets, (, ), one can express 1- 48 with four 3s. Allowing also SQR, one can express 1-64 with four 3s. (iii). Using all 7 symbols in (ii) and brackets, one can express 1- 112 with four 4s. Gives solutions with two and three 4s. (iv). If we also allow $(n) [= Tn, the nth triangle number], he finds solutions with one, two or three 4s. Using log, all integers can be expressed with three 4s.  x  t Three of the best (iii), pp. 17 & 32. Some solutions with one 4, using !, , [n] = INT(n) and $(n). Tony Forbes. Fours. M500 116 (Nov 1989) 4-5. Says someone (possibly Marion Stubbs?) gave a simple variation of King-Hele's and Everett's formulae to use exactly four 4s to yield n. Forbes suggests using one 4 and the three operations: !,  and INT. He already gets stuck at 12. Liz Allen. Brain Sharpeners. Op. cit. in 5.B. 1991. Take two, pp. 96 & 139. Use five 2s to express 1, 2, ..., 26, particularly 17 and 26. P. H. R. Fawlty keys. Mathematical Pie, 129 (Summer 1993) 1023 & Notes, p. 1. Calculator whose only keys are 5, 7, +, , x,  and =. Make numbers from 0 to 20. Solution notes you can make any number by adding enough terms of the form 5  5 and then gives short solutions for 1 through 20. Clifford A. Pickover. Phi in four 4's. Theta (Crewe) 7:2 (Autumn 1993) 58. In Sep 1991 he asked for good approximations to - using four 4s, either with as many symbols as you want or with each symbol used at most four times. Says he was inspired by Conway & Guy's paper of 1962. Brian Boutel produced - = (4 + {4!4})/4. Pickover then extended the question and various solvers got - in five 5s, seven 6s, eight 8s, nine9s and 2k5 ks. John Seldon. Fours. M500 136 (Jan 1994) 15-16. Answers Forbes' 1989 problem of expressing 1 - 100 with one 4 and any number of factorials, x!, and integer square roots, [x]. David Crawford and students. 1999 the end of an era. MiS 28:4 (Sep 1999) 25. Uses 1, 9, 9, 9 to make all integers from 1 to 100. Notes that 2000, 2001, ... are not going to be very useful for such puzzles! Derek Ball. Four 4s. MTg 173 (Dec 2000) 18. Says his fifth year teacher discovered the following for n in terms of four 4s: n = log4/4 log4 n 4, where n denotes nfold repeated square root. Cf Perelman, 1937; KingHele, 1955; Everett, 1971 Everett is very close to this and the others are not quite that close.  7.I.1.LARGEST NUMBER USING FOUR ONES, ETC. Mittenzwey. 1880. Prob. 142, pp. 30 & 7980; 1895?: 162, pp. 34 & 82; 1917: 162, pp. 31 & 7980. Find largest number using four digits. Gets 9^9^9^9 and tries to contemplate its size. 9^9 is given as 387,420,488 (last digit should be 9), so 9^9^9 has 369,693,100 digits. James Joyce. Ulysses. (Dijon, 1922); Modern Library (Random House), NY, 1934, apparently printed 1946. P. 684 (Gardner says the 1961 ed. has p. 699). Bloom estimates that 9^(9^9) would occupy "33 closely printed volumes of 1000 pages each", but he erroneously phrases the number as "the 9th power of the 9th power of 9", which is only 981. King. Best 100. 1927. No. 35, pp. 19 & 46. Largest number using two 4s. Gives 44 = 256. Perelman. FFF. 1934. Four 1's. 1957: prob. 103, pp. 137 & 144145; 1979: prob. 106, pp.167 & 175. = MCBF, prob. 106, pp. 167 & 178 = MCBF, prob. 131, p. 217. "What is the biggest number that can be written with four 1's? Perelman. MCBF. 1937. Probs. 128132, pp. 214219. Largest numbers with: Three twos; Three threes; Three fours; Four ones; Four twos. Sullivan. Unusual. 1947. Prob. 30: Not 999. Largest number that can be written with three integers [sic!]. Answer: 9^(9^9). G. C. S[hephard, ed.] The problems drive. Eureka 11 (Jan 1949) 1011 & 30. No. 5. Find the largest numbers expressible using four 2s or four 4s, no symbols allowed. Answers: 2^2^22; 4^4^4^4. Leroy F. Meyers. An integer construction problem. AMM 66:7 (Aug/Sep 1959) 556561. This deals with Ball's "Four Digits Problem" (see MRE, 6th ed., 1914 in 7.I) and generalizations. In particular, he shows that if one uses 1, 2, 3, 4, with operations +, -, x and brackets, then one can obtain precisely the following: 1, 2, ..., 28, 30, 32, 36. In general he obtains the largest integer expressible using a given multiset of integers (i.e. one is allowed a fixed number of repeats of a value) using the operations +, x and brackets. He also shows that allowing also , for both negation and subtraction, does not increase the maximum obtainable value. He conjectures that allowing also , for both reciprocation and division, does not increase the maximum obtainable, but Meyers has written that a student once showed him a counterexample, but he cannot remember it. He applies his general results to show that no other values are obtainable when using 1, 2, 3, 4. Problematical Recreations 4. Problem 1 and its answer, pp. 3 & 36. (This is one of a series of booklets issued by Litton Industries, Beverly Hills, California, nd [c1963], based on the series of the same name in Aviation Week and Electronic News during 19591971. Unfortunately, neither the date nor location nor author is given and the booklet is unpaginated. The answer simply states the maximum value with no argument.) Reproduced with a proper solution in: Angela Dunn; Mathematical Bafflers; (McGrawHill, 1964, ??NYS); revised and corrected 2nd ed., Dover, 1980, pp. 119 & 132 and with just the answer in: James F. Hurley; Litton's Problematical Recreations; Van Nostrand Reinhold, NY, 1971, chap. 7, prob. 8, pp. 238 & 329. "What is the largest number which can be obtained as the product of positive integers which add up to 100?" (This type of problem must be much older than this?? Meyers writes that he first encountered such problems as an undergraduate in 1947. If one looks at maximizing the LCM instead of the product of the terms, this is the problem of finding a permutation of 100 letters with maximum order.) Sol Golomb. Section 13.8 The minimization of the cost of a digital device (the jukebox problem). IN: Ben Nobel; Applications of undergraduate mathematics in engineering. MAA & Macmillan, 1967, pp. 284286. This considers the problem of Problematical Recreations in the inverse form. We have r kstate devices which allow kr choices and the cost is proportional to rk. Minimize the ratio of cost to capacity or maximize the ratio of capacity to cost. [Another way to express this is to ask which base is best for a computer to use? I recall this formulation from when I was a student in the early 1960s. The answer is e, but here only integer values are used. One can generalise to: given a value, find the smallest sum of numbers whose product is the given value.] The connection with jukeboxes is that they typically have two rows of 12 buttons and one has to press two buttons to make a selection of one from 144 records. One can do much better with five rows of three buttons, but asking a customer to punch five buttons may be unreasonable, so perhaps three rows of five or six buttons might be best. The Fortieth William Lowell Putnam Mathematical Competition, 1 Dec 1979. Problem A1. Reproduced in: Gerald L. Alexanderson, Leonard F. Klosinski & Loren C. Larson; The William Lowell Putnam Mathematical Competition Problems and Solutions: 1965-1984; pp. 33 & 109. "Find positive integers n and a1, a2, ..., an such that a1Ġ+ a2 + ... + an = 1979 and the product a1a2...an is as large as possible." Cliff Pickover & Ken Shirriff. The terrible twos problem. Theta (Crewe) 6:2 (Autumn 1992) 37. They study the problem of making numbers using just +, , x, ^, and 1s (or 1s and 2s). For a given n, what is the least number of digits required? They later permit concatenation, e.g. 11 or 12 is permitted. They report results from various programs and mention some related problems. Bryan Dye. 1, 2, 3, 4 four digits that dwarfed the universe. Micromath 10:3 (Aut 1994) 12-13. Says a version appeared in SA a few years ago and is discussed in: Clifford A. Pickover; Computers and the Imagination; Alan Sutton, 1991. Dye's version is to make the largest number using 1, 2, 3, 4 once and the signs +, , x, , ( ), . (i.e. decimal point). Exponentiation was not considered a sign and was permitted. Pickover's version allowed only the signs , ( ), . (i.e. decimal point). The largest value found actually fits Pickover's conditions: .3^(.2^{.1^4}) has 106990 digits. The largest number using just exponentiation was 2^(3^41)) with 1019 digits. 7.J. SALARY PUZZLE  tIt is better to get a rise of 5 every six months than a rise of 20 every year. The interpretation of the first phrase is somewhat ambiguous see Mills (1993). If the salary is S every six months, the usual interpretation of the first phrase is that the halfyearly payments are: S, S + 5, S + 10, S + 15, S + 20, S + 25, ..., while the second phrase gives payments of: S, S, S + 10, S + 10, S + 20, S + 20, ..., and the former gets 5 extra every year.  t Ball. MRE, 3rd ed., 1896. pp. 26-27. 20 per year versus 5 every half year. He says this is a question "which I have often propounded in past years." It is not in the 1st ed. Workman. Op. cit. in 7.H.1. 1902. Section IX (= Chap. XXXI in c1928 ed.), examples CXLV, prob. 16, pp. 425 & 544 (431 & 577 in c1928 ed.). Compares rise of 15 per year every 3 years with 5 every year. This represents a precursor of the puzzle version. A. Sonnenschein & H. A. Nesbit. The New Science and Art of Arithmetic For the Use of Schools. A. & C. Black, London, 1903. Prob. 12 & 14, pp. 341 & 489. "A youth entered an office at the age of 15 at a salary of 40 a year, with an annual rise of 12. ..." "... What total sum would he have received in 30 years? and what would he have received if the increase had been at the rate of 1 per month?" Mr. X [cf 4.A.1]. His Pages. The Royal Magazine 10:4 (Aug 1903) 336337. Not too obvious. Halfyearly rise of 5 versus 20 a year. No solution given. Susan Cunnington. The Story of Arithmetic. Swan Sonnenschein,, London, 1904. Prob. 35, p. 217. 5 raise each six months versus 10 raise each year. Dudeney?? Breakfast Table Problems No. 330: Smith's salary. Daily Mail (30 & 31 Jan 1905) both p. 7. Raise of 10 per year versus 2 every six months. Pearson. 1907. Part II, no. 87, pp. 132 & 208. As in Ball. Loyd. Salary puzzle. Cyclopedia, 1914, pp. 312 & 381. = MPSL1, prob. 84, pp. 81 & 150-151. = SLAHP: The stenographer's raise, pp. 60 & 108. Raise of 100 per year versus 25 every half year. Interpreting the raise of 25 as worth only 12.50 in a half-year, this option loses, contrary to all other approaches. Clark. Mental Nuts. 1916, no. 6. The two clerks. $25 rise each six months versus $100 rise each year. Dudeney. AM. 1917. Prob. 26: The junior clerk's puzzle, pp. 4 & 150. Two clerks getting 50 per year with one getting a raise of 10 per year versus the other getting a raise of 2 10s every six months with a complicated further process of savings at different rates for five years. M. Adams. Puzzles That Everyone Can Do. 1931. Prob. 276, pp. 103 & 171: A problem in salaries. 20 rise every six months versus 80 rise each year. T. O'Conor Sloane. Rapid Arithmetic. Quick and Special Methods in Arithmetical Calculation Together with a Collection of Puzzles and Curiosities of Numbers. Van Nostrand, 1922. [Combined into: T. O'Conor Sloane, J. E. Thompson & H. E. Licks; Speed and Fun With Figures; Van Nostrand, 1939.] The two clerks, p. 168. Raise of $50 every six months versus $200 each year. F. & V. Meynell. The Week-End Book, op. cit. in 7.E. 1924. Prob. one, p. 274 (2nd ed.), pp.406-407 (5th? ed.). Raise of 20 per year versus 5 every half year. Peano. Giochi. 1924. Prob. 16, p. 5. 1000 per year with rise of 20 each year versus 500 each halfyear with rise of 5 each halfyear. Wood. Oddities. 1927. Prob. 31: A matter of incomes, p. 31. $1000 per year with $20 per year increase versus "$5 each half year increase". Collins. Fun with Figures. 1928. Do figures really lie?, pp. 3536. $50 every six months versus $200 per year. R. Ripley. Believe It Or Not! Book 2. (Simon & Schuster, 1931); Pocket Books, NY, 1948. P. 123: Figure your raise in pay. A raise of one every day is better than a raise of 35 every week. (Assumes a six day week.) Phillips. Week-End. 1932. Time tests of intelligence, no. 20, pp. 16 & 189-190. Raise of 2000 per year versus 500 half-yearly. Phillips. Brush. 1936. Prob. G.1: The two clerks, pp. 20 & 87. Raise of 200 annually versus 50 half-yearly. McKay. At Home Tonight. 1940. Prob. 4: A choice of rises, pp. 63 & 76. 5 per halfyear versus 15 per year. Solution is unclear and seems to be wrong. " 5 each six months is 5 in the first halfyear and 10 in the second that is, 15 per year. But the man who gets 5 per six months gets 5 in the first year, and of course he keeps this advantage year by year." I get that the first case is ahead by 5n in the nth year. Sullivan. Unusual. 1943. Prob. 5: Raising the raise question. Raise of $20 per year versus $5 every half year. Birtwistle. Calculator Puzzle Book. 1978. Prob. 56: The interview, pp. 39 & 99100. 3050 yearly plus 100 each year versus 1500 halfyearly plus 50 each halfyear. D. J. Hancox, D. J. Number Puzzles For all The Family. Stanley Thornes, London, 1978. Puzzle 11, pp. 4 & 48.. 1 rise per month every month versus 144 rise per year every year. Says the first gives 66 extra in the first year and 210 extra in the second year, while the second gives no extra in the first years and 144 extra in the second year. He then says: "Hence he would never get the 66 he would have received in the first year." In fact, he loses 66 every year. Stuart E. Mills. Dollars and sense. CMJ 24 (1993) 446448. Raise of $1000 per year versus $300 every half year. Discusses various interpretations of the second phrase and gives some recent references. Comments by myself and various others seem to have appeared (??NYS) as they are included in the collection of these columns: Edward J. Barbeau; Mathematical Fallacies, Flaws, and Flimflam; Spectrum Series, MAA, 2000, pp. 1114. This gives references to various recent appearances of the problem in 1943, 1983, 1992. John P. Ashley. Arithmetickle. Arithmetic Curiosities, Challenges, Games and Groaners for all Ages. Keystone Agencies, Radnor, Ohio, 1997. P. 80: The best job offer. $20,000 per year plus a $500 raise every six months versus $20,000 per year plus a raise of $1000 each year. He says the sixmonthly payments of the first are 10,000, 10,500, 11,000, 11,500, ... and the payments for the second are 10,000, 10,000, 11,000, 11,000, .... But the second is a raise of $2,000 per year! 7.K.CONGRUENCES The Friday Night Book (A Jewish Miscellany). Soncino Press, London, 1933. Mathematical Problems in the Talmud: The Divisibility Test, p. 137. Hebrew law requires fields to lie fallow every seventh year, and this is to hold for all fields at once! Rabbi Huna gave the following rule. Write the year as y = 100a + b. Form 2a + b. Then the year is a Sabbatical year if 7 divides 2a + b. No explanation is given in the Talmud, but we clearly have y  2a + b (mod 7). [The Talmud was compiled in the period 300 to 500. This source says Rabbi Huna is one of the few mathematicians mentioned in the Talmud, but gives no dates and he is not mentioned in the EB. From the text of another problem attributed to him (cf in Section 6.AD), the problem would seem to be sometime in the 15 C.] 7.K.1.CASTING OUT NINES  tSee Smith, History II 151-154 for a detailed discussion. He says it appears in al-Khowarizmi and al-Karkhi and that it is generally assumed to come from India, but his earliest Indian source is Lilavati, 1150. G. R. Kaye; References to Indian mathematics in certain Medi%val works; J. Asiatic Society of Bengal (NS) 7:11 (Dec 1911) 801816 notes the appearances in alKhwrizm3, Avicenna and Maximus Planudes [Arithmetic after the Indian method; c1300; op. cit. in 7.E.1] but asserts it does not occur in early Indian sources but cf Aryabhata II, 950. Dickson I, chap. XII, pp. 337-346, especially p. 337, gives a concise history. He says al-Karkhi was the first to use a (mod 11) check. See Tropfke, pp. 165167. I have recently realised that certain puzzle problems should be listed here, but so far I have only noted Boy's Own Book, Boy's Own Book (Paris), Carroll, Peano, Parlour Games for Everyone there must be many more 19C and even 18C examples. Basically these involve getting someone to produce a number divisible by nine and asking him to delete one digit and tell you the others you tell him the missing digit. There are many of these and I probably won't try to record all of them, but subtracting a number from its reversal may be the forerunner of the 1089 puzzle of 7.AR.  tSt. Hippolytus. ) %3 # % 3  /% (??= Philosphumena) (= Refutatio Omnium Haeresium = Refutation of all Heresies). c200. Part iv, c. 14. ??NYS. Discusses adding up digits corresponding to letters (mod 9) and mentions considering it (mod 7). (HGM I 115-117) See also: Smith II 152, Dickson I 337 and Saidan (below), p.472. Hippolytus doesn't use the method to check any arithmetic. (St. Hippolytus may be the only antipope to be counted a saint! A reference says he was Bishop of Portus and the MS was discovered at Mt. Athos in 1842.) Iamblichus. On Nicomachus's Introduction to Arithmetic. c325. ??NYS. In: SIHGMI108-109. Special case. (See also HGM I 114-117.) Muhammad ibn MEs al-Khwrizm3. c820. Untitled Latin MS of 13C known as Algorismus or Arithmetic, Cambridge Univ. Lib. MS Ii.6.5. Facsimile ed., with transcription and commentary by Kurt Vogel as: Mohammed ibn Musa Alchwarizmi's Algorismus, Das frGheste Lehrbuch zum Rechnen mit indischen Ziffern; Otto Zeller Verlagsbuchhandlungen, Aalen, 1963. English translation by John N. Crossley & AlanS. Henry as: Thus spake al-Khw]rizm: A translation of the text of Cambridge University Library Ms.Ii.vi.5; HM 17 (1990) 103-131. [Crossley & Henry name this author Ab Jafar Muhammad ibn Ms] al-Khw]rizm, but I have seen no other authority giving Ab Jafar several give Ab Abdall]h and Rosen's translation The Algebra of Mohammed ben Musa specifically says our author "must therefore be distinguished from Abu Jafar Mohammed ben Musa, likewise a mathematician and astronomer, who flourished under the Caliph Al Motaded" (c900). F. 108r = Vogel p.25 = Crossley & Henry p. 117. Describes casting out 9s in doubling and in multiplication. Aryabhata II. Mah-siddhnta. 950. Edited by M. S. Dvivedi, Braj Bhushan Das & Co., Benares, 1910. English Introduction, pp. 21-23; Sanskrit text, p. 245. Casting out 9s for multiplication, division, squaring, cubing, and taking square and cube roots. (Datta & Singh I 181 give the text in English.) AbE alHassan Ahmad Ibn Ibrh3m alUql3dis3. Kitb al FusEl f3 al-Hisb al-Hind3. 952/953. MS 802, Yeni Cami, Istanbul. Translated and annotated by A. S. Saidan as: The Arithmetic of Al-Uqldis; Reidel, 1978. Book II, chap. 13, pp. 153-155 and Book III, chap. 7-8, pp. 195-201 deal with checking by casting out 9s, which is given only briefly, apparently being well-known. He applies it to division and square roots. The method is also mentioned in Book II, chap. 2. On pp. 468-472, Saidan discusses the appearance of various rules in early texts. His earliest Indian example is Lilavati, 1150, but he gives no reference. Kshy]r ibn Labb]n = Ab alHasan Ky]r ibn Labb]n ibn B]ahri alvl. Kit]b f usl His]b alHind [Principles of Hindu Reckoning]. c1000. Facsimile with translation by Martin Levey & Marvin Petruck. Univ. of Wisconsin Press, Madison, 1965.  x  thFirst Book, Ninth Section: On arithmetic checks, f. 274a, pp. 7071. Brief description of casting out nines. Second Book, Eleventh Section: On checks, ff. 280a280b, pp. 9497. Brief description of casting out nines in base 60. Introduction, pp. 3233, discusses the above, noting that use of 9 in base 60 is unreasonable, but others also did it. Says Sibt alM]ridn (16C) used casting out 8s and 7s in base 60, but that alK]sh (15C) used casting out 59s in base 60. Kshy]r does not state that the check proves the correctness of the result, though this was commonly believed, e.g. by Fibonacci and Sibt, though alK]sh clearly discusses the question.  x  tIbn Sina = Avicenna. Treatise on Arithmetic. c1020. ??NYS. Complete rules for checking operations by casting out 9s, attributed to the Hindus, (Smith, Isis 6 (1924) 319). (See also Cammann 3 (cited in 7.N); Datta & Singh, I, 184; and Kaye, above, who cite F.Woepcke; M)moire sur la propagation des chiffres indiens; J. Asiatique (6) 1 (1863) 27529; p. 502, ??NYS.) The DSB entry indicates that the material is in Ibn Sina's Al-Shif' (The Healing) and there doesn't appear to be a translation. Suter 89 mentions some Latin translations but I'm not clear whether they are this book or a related book. Saidan's discussion says Woepcke (p. 550 [sic]) construes ibn Sina as saying that the method is Indian, but this is a contentious interpretation. Kaye, above, says Woepcke is wrong. Smith, History II 151, says the expression "has been variously interpreted". Bhaskara II. Lilivati. 1150. Smith, History II 152, cites this in Taylor's edition, p. 7, but the method is not in Colebrooke and neither Dickson nor Datta & Singh cite it, so perhaps it is an addition in the text Taylor used?? Fibonacci. 1202. Pp. 89, 20, 39, 45 (S: 2426, 41, 67, 74) uses checks (mod 7, 9 and 11). On p. 8 (S: 24), he implies that if the 'proof' is right, then the calculation is correct see comments at Kshy]r above. Maximus Planudes. 0--# )'  +%       (Psephephoria kat' Indous e Legomene Megale) (Arithmetic after the Indian method). c1300. (Greek ed. by Gerhardt, Das Rechenbuch des Maximus Planudes, Halle, 1865, ??NYS [Allard, below, pp. 2022, says this is not very good]. German trans. by H. Waeschke, Halle, 1878, ??NYS [See HGM II 549; not mentioned by Allard].) Greek ed., with French translation by A. Allard; Maxime Planude Le Grand Calcul selon des Indiens; Travaux de la Facult) de Philosophie et Lettres de l'Univ. Cath. de Louvain XXVII, Louvain-la-Neuve, 1981. Proofs by casting out 9s are given in the material on the operations of arithmetic. Narayana Pandita (= N]r]yana Pandita). Ganita-kaumud. 1356. Edited by P. Dvivedi, Indian Press, Benares, 1942. Introduction in English, p. xv, discusses the material. Allows any modulus. (English in Datta & Singh I 183.) The Treviso Arithmetic = Larte de labbacho. Op. cit. in 7.H. 1478. F. 4v onward (p. 46 in Swetz) uses casting out 9s as a check on many examples. Swetz (p.189) refers to Avicenna and the Hindus. On f. 10v (Swetz p.59), the anonymous author says that proving a subtraction by addition "is more rapid and also more certain than the proofs by 9s" and he makes similar statements regarding multiplication and division. On p. 323 of his Isis article, Smith says the author "gives a proof by casting out sevens". This would be on or near f. 17v (Swetz p.73). I can find nothing of the sort the author has an example of multiplication by 7, but he checks it by casting out 9s. Borghi. Arithmetica. 1484. Ff. 8r9r (1509: ff. 9r9v). Casting out 7s and 9s. This is applied over the next few sections, but I don't see any indication that casting out 9s is not a certain test. However he uses casting out 7s more often than 9s which may indicate that he was aware that 7s is a more secure test than 9s. Chuquet. 1484. Triparty, part 1. English in FHM 4142. "There are several kinds of proofs such as the proof by 9, by 8, by 7, and so on by other individual figures down to 2, .... Of these only the proof by 9, because it is easy to do, and the proof by 7, because it is even more certain than that by 9 are treated here." He then notes that these proofs are not always certain. Pacioli. Summa. 1494. Ff. 20v23v. Discusses casting out 9s and 7s and notes that these tests are not sufficient. Apianus. Kauffmanss Rechnung. 1527. Gives numerous examples of testing by 9s, and also by 8s, 7s and 6s, in his sections on the four arithmetic operations and also under arithmetic progressions. Recorde. First Part. 1543. Discusses the proof by nines in his chapters on: addition, ff. D.i.r D.iii.v (1668: 2932: The proof of Addition); subtraction, ff. F.iii.r F.iiii.r (omitted in 1668); multiplication, ff. G.vi.r G.vi.v (1668: 7072: Proof of Multiplication); and division, ff. H.iii.v H.iiii.v (1668: 8284: Proof of Division). Hutton. A Course of Mathematics. 1798? 1833 & 1857: 612. In his discussion of the basic arithmetic operations, we find on p. 7 under To Prove Addition, "Then, if the excess of 9's in this sum, ...., be equal to the excess of 9's in the total sum ..., the work is right." A footnote explains the idea and is less clear as to the direction of implication being asserted: "it is plain that this last excess must be equal to the excess of 9's contained in the total sum". The note concludes: "This rule was first given by Dr Wallis in his Arithmetic, published in the year 1657." However, Hutton does not mention the rule under subtraction and under multiplication on pp. 1011, he says the "remainders must be equal when the work is right." All in all, it seems that he is surprisingly unclear for his time. Boy's Own Book.  x  thThe number nine: "To add a figure to any given Number, which shall render it divisible by Nine". 1828: 179180; 18282: 235; 1829 (US) & 1881 (NY): 103; 1855&1859:389; 1868:429; 1880: 459. Here the digit is actually added, but then he indicates that it can be inserted. [Incidentally, this avoids the hazard of discovering the missing digit might be either 0 or 9.] This section is extended and combined into Properties of certain numbers from 1868. Cf 1843(Paris):342 The cancelled figure guessed. 1828: 177; 18282: 237; 1829 (US) & 1881 (NY): 105; 1843(Paris): 346: A person striking a figure out of the sum of two given numbers, to tell what that figure was; 1855&1859:392393; 1868: 430431; 1880:460-461. Give a person several multiples of nine, then ask him to add two of them and strike out one digit from the total and tell you the sum of the other digits in the total. No consideration of the case when the cancelled digit could be 0 or 9. = Boy's Treasury, 1844, pp. 303304. = de Savigny, 1846, p. 346.  x  tBoy's Own Book. 1843 (Paris): 342. "To make any number divisible by nine, by adding a figure to it." Only appends or inserts the necessary digit. = Boy's Treasury, 1844, p.299. Lewis Carroll. Diary entry for 8 Feb 1856, In CarrollGardner, pp. 4344. Observes that a number minus its reverse is divisible by nine, so you can ask someone to delete a digit and show you the rest and you announce the deleted digit. Gardner points out that one can subtract any permutation of the original digits. Mittenzwey. 1880.  x  thProb. 4041, pp. 9 & 6061; 1895?: 4647, pp. 14 & 64; 1917: 4647, pp.13 & 5859. Deduce the figure deleted in 9x; in x minus the sum of its digits. 1895?: prob. 118, pp. 7475; 1917: 118, pp. 23 & 7172. Casting out 9s as a method of checking multiplication, claiming that the verification shows the calculation is correct.  x  tParlour Games for Everybody. John Leng, Dundee & London, nd [1903 BLC], p. 42: The expunged figure. Have someone write a number, form the sum of its digits and subtract that from the given number. Get him to strike out any figure and tell you the sum of the remaining figures. Says that if the result is a multiple of nine, then a nine was struck out. Peano. Giochi. 1924. Prob. 50, p. 13. Take x, form 10x and subtract x. Cancel a nonzero figure from the result and tell me the other figures. I will tell what number you cancelled. 7.L.GEOMETRIC PROGRESSIONS  tSee Tropfke 628. These also occur in 7.M, 7.S.1 and 10.A. I am starting to include early problems which involve interpolation in a geometric series here these were normally solved by linear interpolation. From about 1400, such problems arise in compound interest but I will omit most such problems. See Chuquet here and Chiu Chang Suan Ching & Cardan in 10.A.  t H. V. Hilprecht. Mathematical, Metrological and Chronological Tablets from the Temple Library of Nippur. Univ. of Pennsylvania, Philadelphia, 1906. Pp. 13, 28-34, 62, 69, pl.15, PL. IX, are about a tablet which has a geometric progression from c-2300. The progression is double: an = 125 * 2n and 604/an for n=0,1,..., 7. There is no summation. Tablet K 90 of the British Museum. A moon tablet deciphered by Hincks containing 5,10,20,40, 80 followed by 80, 96, 112, 128, ..., 240. Described in The Literary Gazette (5 Aug 1854) ??NYS. Cited and described in: Nicomachus of Gerasa: Introduction to Arithmetic; Translated by Martin Luther D'Ooge, with notes by Frank Egleston Robbins and Louis Charles Karpinski; Macmillan, London, 1926; p. 12. Euclid. IX: 35, 36. This gives the general rule for the sum of a geometric progression. The Friday Night Book (A Jewish Miscellany). Soncino Press, London, 1933. Mathematical Problems in the Talmud, pp. 132133. The Talmud says that any one visiting a sick person takes away a sixtieth of his illness. This led to the question of what happened if sixty people visited the person. This was answered by saying that the visitor took away a sixtieth of the illness that the person had, i.e. the patient was left with 59/60 of his illness, so that 60 visits left him still with (59/60)60 of his illness. The text quoted in the source says this 'is still approximately onequarter of the original illness', but it is .36479. The modern compiler adds that 'The Talmud does not indicate the method of working out the remainder after each visitor, and it is to be noted that although the summation of series was known to the Greeks, there is no mention of it anywhere in the Talmud.' To me, this shows some confusion as I don't see that summation of series is needed! [The Talmud was compiled in the period 300 to 500, but nothing in the source gives any more precise dating for this problem.] Zhang Qiujian . Zhang Qiujian Suan Jing. Op. cit. in 7.E. 468, ??NYS. Mikami 42 gives: "A horse, halving its speed every day, runs 700 miles in 7 days. What are his daily journeys?" i.e. x*(1+1/2+... + 1/64) = 700. Solved by adding up. Mahavira. 850.  x  thChap. IV, v. 28, p. 74: x - x/2 - x/4 - ... - x/256 = 32. Chap. VI, v. 314, pp. 175-176: Let ai+1 = r*ai + c. He sums such terms.  x  tFibonacci. 1202. Pp. 313-316 (S: 439443). Man has 100 and gives away 1/10 of his wealth 12 times. This has been described under 7.E. Lucca 1754. c1330. F. 10v, pp. 36-37. Computes 240 & 2100 by repeated squaring. Columbia Algorism. c1350. Prob. 63, pp. 84-85. Same as the Fibonacci, but he converts to pounds, shillings and pence! Folkerts. Aufgabensammlungen. 1315C. Four sources with progressions with ratio 7 and seven sources with ratio 12. Chuquet. 1484. He gives a number of such problems see also 7.E.  x  thProb. 96, English in FHM 219. Cask of 9 drains so first barrel takes 1 hour, second barrel takes 2 hours, third barrel takes 4 hours, .... How long to empty? I.e. he wants the sum of 9 terms of a geometric progression. He gets the correct answer of 29.5 1 hours. Prob. 97, English in FHM 219. Man travels 1, 3, 9, ... leagues per day. How far has he travelled in 5 days? He gets the correct answer of (35.5ĩ1)/2 days.  x  tGhaligai. Practica D'Arithmetica. 1521. Prob. 29, f. 66v. Same as Fibonacci. (H&S 5960.) Buteo. Logistica. 1559.  x  thProb. 69, pp. 276278. 1 + .9 + (.9)2 + ... + (.9)x = 7.5. The phrasing of the problem is unclear, but this is what he considers. He interpolates linearly between 12 and 13, getting 12.164705107, while the exact answer is (log.25/log.9)-1=12.15762696. Prob. 84, pp. 294296. Relates 2i and 5i for i = 1, ..., 7. He determines 23.5 as 128 which he estimates as 11@. Similarly he estimates 53.5 as 279.  x  tJacques Chauvet Champenois. Les Institutions de L'Arithmetique. Hierosme de Marnef, Paris, 1578, p. 70. ??NYS. Problem of tailor and robe involving 4888 divided by 2 twenty times. (French quoted in H&S 14-15.) van Etten. 1624. Prob. 87 (84): Des Progressions & de la prodigieuse multiplication des animaux, des Plantes, des fruicts de l'or & de l'argent quand on va tousjours augmentant par certaine proportion, pp. 111-118 (177-183). Numerous examples including horseshoe problem and chessboard problem, with ratios 1000, 4, 50. Henrion's Notte, p.38, observes that there are many arithmetical errors which the reader can easily correct. In part X: Multiplication des Hommes, he considers one of the children of Noah, says a generation takes 30 years and that, when augmented to the seventh, one family can easily produce 800,000 souls. The 1674 English ed. has: "... if we take but one of the Children of Noah, and suppose that a new Generation of People begin at every 30 years, and that it be continued to the Seventh Generation, which is 200 years; ... then of one only Family there would be produced 111000 Souls, 305 to begin the World: ... which number springing onely from a simple production of one yearly ...." W. Leybourn. Pleasure with Profit. 1694. Chap. VI, pp. 2428: Of the Increase of Swine, Corn, Sheep, &c. Examples with ratios 4, 40, 2, 1000, 2, mostly taken from van Etten. Then art. VI: Of Men, discusses the repopulation of the world from Noah's children: "... if we take but one of the Children of Noah, and suppose that a New Generation of People begin at every 30 years, and that it be continued to the seventh Generation, which is 210 years; ... then, of one only family there would be produced 111305, that is, One hundred and eleven thousand, three hundred and five Souls to begin the World .... ... such a number arising only from a simple production of only One yearly ...." I cannot work out how 111305 arises the fact that he spells it out makes it unlikely to be a misprint. Ozanam. 1694. Prob. 8, 1696: 3335; 1708: 2932. Prob. 11, 1725: 6875. Section II, 1778: 6874; 1803: 7076; 1814: ??NYS; 1840: 3436. A discussion of geometric progression and a mention of 1,2,4, .... 1778 et seq. also mention 1, 3, 9, .... Ozanam. 1725. Prob. 11, questions 6 & 7, 1725: 79-82. Prob. 3, parts 13, 1778: 8082; 1803: 8284; 1814: 7275; 1840: 3839. Examples of population growth in Biblical and biological contexts. In 1725, he has ratios of 2, 50, 3, 4, 1000, The examples vary a bit between 1725 and 1778. Walkingame. Tutor's Assistant. 1751. The section Geometrical Progression gives several problems with powers of 2 and the following less common types.  x  thProb. 5, 1777: p. 95; 1835: p. 103; 1860: p. 123. Find 1+4+ 16 + ... + 411 farthings. Prob. 8, 1777: p. 96; 1835: p. 104; 1860: p. 123. Find 2+6+ 18 + ... + 2 x 321. If these are pins, worth 100 to the farthing, what is the value?  x  tVyse. Tutor's Guide. 1771? The section Geometrical Progression, 1793: 35, pp. 138143; 1799: XXXV, pp. 146151 & Key pp.190192, gives several examples with doublings and triplings as well as examples with ratios of 3/2 and 10. There is a major error in the solution of prob. 7, to find 2+6+ 18 + ... + 2 x 319. Pike. Arithmetic. 1788. Pp. 237239. Numerous fairly standard examples, mostly doubling, but with examples of powers of 3 and of 10 and the following. D. Adams, 1835, copies two examples, but not the following.  x  thPp. 239240, no. 8. One farthing placed at 6% compound interest in year 0 is worth what after 1784 years? And supposing a cubic inch of gold is worth 53 2s 8d, how much gold does this make? This is very close to 2150 farthings and makes about 4 x 1014 solid gold spheres the size of the earth!  x  tEadon. Repository. 1794. P. 241, ex. 3. Doubling 20 times from a farthing. John King, ed. John King 1795 Arithmetical Book. Published by the editor, who is the greatgreatgrandson of the 1795 writer, Twickenham, 1995. P. 100. 10+102Ġ+...+1011 grains of wheat, converted to bushels and value at 4s per bushel. (Beeton's) Boy's Own Magazine 3:6 (Jun 1889) 255 & 3:8 (Aug 1889) 351. (This is undoubtedly reprinted from Boy's Own Magazine 1 (1863).) Mathematical question 59. Seller of 12 acres asks 1 farthing for the first acre, 4 for the second acre, 16 for the third acre, .... Buyer offers 100 for the first acre, 150 for the second acre, 200 for the third acre, .... What is the difference in the prices asked and offered? Also entered in 7.AF. Lewis Carroll. Sylvie and Bruno Concluded. Macmillan, London, 1893. Chap. 10, pp.131-132. Discusses repeated doubling of a debt each year as a way of avoiding paying the debt "You see it's always worth while waiting another year, to get twice as much money!" =CarrollWakeling II, prob. 5: A new way to pay old debts, pp. 9 & 66, where Wakeling adds some problems based on repeated doubling and gives the chessboard problem. 7.L.1.1 + 7 + 49 + ... & ST. IVES See Tropfke 629. Papyrus Rhind, c-1650, loc. cit. in 7.C. Problem 79, p. 112 of vol. 1 (1927) (= p. 59 (1978)). 7 + 49 + 343 + 2401 + 16807. (Sanford 210 and H&S 55 give Peet's English.) Houses, cats, mice, ears of spelt, hekats. L. Rodet. Les pr)tendus probl/mes d'alg/bre du manuel du calculateur (gyptien (Papyrus Rhind). J. Asiatique 18 (1881) 390-459. Appendice, pp. 450-454. Discusses this problem and its appearance in Fibonacci (below). F. Cajori. History of Mathematics. 2nd ed., Macmillan, 1919; Chelsea, 1980. P. 90 gives the legend that Buddha was once asked to compute 717. Shakuntala Devi. The Book of Numbers. Orient Paperbacks (Vision Books), Delhi, 1984. This gives more details of the Buddha story, saying it occurs in the Lalitavistara and Buddha finds the number of atoms (of which there are seven to a grain of dust) in a mile, obtaining a number of 'about 50 digits'. Note: 758 = 1.04 x 1049. Alcuin. 9C. Prob. 41: Propositio de sode et scrofa. This has sows which produce 7 piglets, but this results in a GP of ratio 8. Fibonacci. 1202.  x  thPp. 311-312 (S: 438). 7 + ... + 117649. Old women [The Latin is vetule, which is corrupt and the gender is not clear M Sigler says old men.], mules, sacks, loaves, knives, sheathes. (English in: N. L. Biggs; The roots of combinatorics; HM 6 (1979) 109-136 (on p. 110) and Sanford 210. I have slides of this from L.IV.20 & 21. It is on f. 147r of L.IV.20 and f. 225r of L.IV.21.) P. 312 (S: 438439). 100 + 10000 + ... + 108. Branches, nests, eggs, birds.  x  tMunich 14684. 14C. Prob. XXXI, pp. 83-84. Refers to 7 + 49 + ... + 117649. AR. On p. 227, Vogel refers to an example in CLM 4390 which has not been published. Peter van Halle. MS. 3552 in Royal Library Brussels, beginning "Dit woort Arithmetica coomt uuter griexer spraeken ...." 1568. F. 23v. "There were 5 women and each woman had 5 bags but in each bag were 5 cats and each cat had 5 kittens question how many feet are there to jump with?" Copy of original Dutch text and English translation provided by Marjolein Kool, who notes that van Halle only counts the feet on the kittens. Josse Verniers. MS. 684 in University Library of Ghent, beginning "Numeration heet tellen ende leert hoemen die ghetalen uutghespreken ende schrijven ...." 1584. F. 7v. "Item there is a house with 14 rooms and in each room are 14 beds and each bed lay 14 soldiers and each soldier has 14 pistols and in each pistol are 3 bullets Question when they fire how many men, pistols and bullets are there" Copy of original Dutch text and English translation provided by Marjolein Kool. Harley MS 7316, in the BM. c1730. ??NYS quoted in: Iona & Peter Opie; The Oxford Dictionary of Nursery Rhymes; OUP, (1951); 2nd ed., 1952, No. 462, p. 377. The Opies give the usual version with 7s, but their notes quote Harley MS 7316 as: "As I went to St. Ives I met Nine Wives And every Wife had nine Sacs And every Sac had nine Cats And every Cat had Nine Kittens." The Opies' notes also cite Mother Goose's Quarto (Boston, USA, c1825), a German version with 9s and a Pennsylvania Dutch version with 7s. Halliwell, James Orchard. Popular Rhymes & Nursery Tales of England. John Russell Smith, London, 1849. Variously reprinted my copy is Bodley Head, London, 1970. P. 19 refers to "As I was going to St. Ives" in MS. Harl. 7316 of early 18C, but doesn't give any more details. D. Adams. Scholar's Arithmetic. 1801. As I was going to St. Ives, I met seven wives, Every wife had seven sacks, Every sack had seven cats, Every cat had seven kits, Kits, cats, sacks and wives, How many were going to St. Ives? No solution. Child. Girl's Own Book. 1842: Enigma 35, pp. 233234; 1876: Enigma 27, pp. 196197. "As I was going to St. Ives, I chanced to meet with nine old wives: Each wife had nine sacks, Each sack had nine cats, Each cat had nine kits; Kits, cats, sacks, and wives, Tell me how many were going to St. Ives?" Answer is "Only myself. As I met all the others, they of course were coming from St. Ives." The 1876 has a few punctuation changes. = Fireside Amusements. 1850: No. 48, pp. 114 & 181; 1890: No. 34, p. 102. The 1850 solution is a little different: "Only myself. As I was going to St. Ives, of course all the others were coming from it." The 1890 solution differs a little more: "Only myself. As I was going to St. Ives, all the others I met were coming from it." Kamp. Op. cit. in 5.B. 1877. No. 20, p. 327. 12 women, each with 12 sticks, each with 12 strings, each with 12 bags, each with 12 boxes, each with 12 shillings. How many shillings? Mittenzwey. 1880. Prob. 20, pp. 3 & 59; 1895?: 24, pp. 9 & 63; 1917: 24, pp. 9 & 57. Man going to St?tteritz meets 9 old women, each with 9 sacks, each with 9 cats, each with 9 kittens. How many were going to St?tteritz? Answer is one. Heinrich Voggenreiter. Deutsches Spielbuch Sechster Teil: Heimspiele. Ludwig Voggenreiter, Potsdam, 1930. Pp. 105106: Wieviel FGsse sind es? Hunter going into the woods meets an old woman with a sack which has six cats, each of which has six young. How many feet all together were going into the woods? His answer is 'Only one', which has confused 'feet' with 'walker' this may be an obscure German usage, but I can't find it in my dictionaries. Joseph Leeming. Riddles, Riddles, Riddles. Franklin Watts, 1953; Fawcett Gold Medal, 1967. P. 150, no. 11: "As I was going to St. Ives, I chanced to meet with nine old wives; ...." [I don't recall any other contemporary examples using 9 as multiplier.] Mary & Herbert Knapp. One Potato, Two Potato ... The Secret Education of American Children. Norton, NY, 1976. Pp. 107108 gives a modern New York City version: "There once was a man going to St. Ives Place. He had seven wives; each wife had seven sacks; each sack had seven cats; each cat had seven kits. How many altogether were going to St. Ives Street? One." St. Ives has become attached to a location in New York! Colin Gumbrell. Puzzler's A to Z. Puffin, 1989. Pp. 9 & 119: As I was going ... "As I was going to St Ives, I met a man with seven wives; And every wife had seven sons; But they were not the only ones, For every son had seven sisters! Bewildered by so many misters And by so many misses too, I quickly cried: 'Bonjour! Adieu!' And hurried to another street, Away from all their trampling feet. Now, here's the point that puzzles me yet: Just how many people had I met?" Answer is 106, or 64 if there are just seven girls who are halfsisters to all the 49 sons. About 2000, someone told me that the answer to the classic St. Ives riddle ought to me 'None' as it asks how many of the kits, cats, sacks, and wives were going to St. Ives. Ed Pegg Jr, Marek Penszko and Michael Kleber circulated a new version in early 2001 on the Internet. Tim Rowett has adapted one of the St Ives postcards with this new text. As I was going to St Ives I met a man with seven wives Every wife had seven sacks Every sack had seven cats Every cat had seven kits We traded bits Each cat, sack, wife, and he Took a kit. The rest for me. So now I have a kit supply. How many kits did I just buy? The St. Ives of the riddle is usually thought to be the one in Cornwall, but there are also St. Ives in Cambridgeshire (near Huntingdon) and in Dorset (near Ringwood) and a St. Ive in eastern Cornwall (near Liskeard). Darrell Bates. The Companion Guide to Devon and Cornwall. Collins, 1976. P. 301 says the Cornish St. Ives is named for a 6C Irish lady missionary named St. Ia who crossed the Irish Sea on a leaf. John Dodgson [Home Town What's behind the name; Drive Publications for the Automobile Association, Basingstoke, 1984; p. 40] agrees. Gilbert H. Doble. St. Ives Its Patron Saint and its Church. Cornish Parish Histories No. 4, James Lanham Ltd, St. Ives, 1939. This says the lady was named Ya, Hya or Ia. The 'v' was probably inserted due to the influence of the Breton St. Yves, with the first appearance of the form 'Ives' being in 1571. The earliest reference to St. Hya is c1300 and says she was an Irish virgin of noble birth, who found her friends had departed for Cornwall. As she prayed she saw a leaf in the water and touched, whereupon it grew big enough to support her and she was wafted to Cornwall, arriving before her friends. Doble suggests that Ireland refers to Wales here. The next mention is in 1478 and says she was the sister of St. Erth and St. Uny and was buried at St. Hy. There is only one other old source, a mention in 1538. There seems to be very little, if anything, known about this saint! The Michelin Green Guide to Brittany (3rd ed., Michelin et Cie, Clermont-Ferrand, 1995, pp.178 & 237238) describes the Breton St. Yves, which I had assumed to be the eponym of the Cornish St. Ives. St. Yves (Yves Helori (1253-1303)) was once parish priest at Louannec, C=tesd'Armor, where a chasuble of his is preserved in the church. His tomb is in the Cathedral of St. Tugdual in Tr)guier, C=tesd'Armor. His head is in a reliquary in the Treasury. He worked as a lawyer and is the patron saint of lawyers. He was born in the nearby village of MinihyTr)guier and his will is preserved in the Chapel there. The Chapel cemetery contains a monument known as the tomb of St. Yves, but this is unlikely to contain him. Attending his festival, known as a 'pardon', is locally known as 'going to St. Yves' !! The Cornish and Breton stories may have influenced each other. 7.L.2.1 + 2 + 4 + ... See HQyrup in 7.L.2.a for other early examples of doubling 30 times. Chiu Chang Suan Ching. c-150? Chap. III, prob. 4, pp. 28-29. Weaver weaves a(1+2+4+8+16), making 5 in all. (English in Needham, pp.137-138. Needham says this problem also occurs in Sun Tzu (presumably the work cited in 7.P.2, 4C), ??NYS.) Alcuin. 9C. Prob. 13: Propostio de rege et de ejus exercitu. 1+1+2 + 4 + ... + 229 = 230. Calculations are suppressed in the Alcuin text, but given in the Bede. Murray 167 wonders if there is any connection between this and the Chessboard Problem (7.L.2.a). Bhaskara II. Lilavati. 1150. Chap. V, sect. II, v. 128. In Colebrooke, pp. 55-56. 2+4+...+230. W. Leybourn. Pleasure with Profit. 1694. See in 7.L. Wells. 1698. No. 103, p. 205. Weekly salary doubles each week for a year: 1+2+4+...+251. Ozanam. 1725. Prob. 11, question 5, 1725: 78-79. 1 + 2 + ... + 231. Walkingame. Tutor's Assistant. 1751. The section Geometrical Progression gives several problems with straightforward doublings see 7.L and 7.L.2.b for some more interesting examples. Vyse. Tutor's Guide. 1771? Same note as for Walkingame. Eadon. Repository. 1794. P. 241, ex. 3. Doubling 20 times from a farthing. John King, ed. John King 1795 Arithmetical Book. Published by the editor, who is the greatgreatgrandson of the 1795 writer, Twickenham, 1995. P. 115. For 20 horses, is starting with a farthing and doubling up through the 19th horse, with the 20th free, more or less expensive than 20 per horse? Boy's Own Book.  x  thCurious calculation. 1868: 433. 1 + 2 + ... + 251 pins would weigh 628,292,358 tons and require 27,924 ships as large as the Great Eastern to carry them. Arithmetical [sic] progression. 1868: 433. 1 + 2 + ... + 299 farthings. Answer is wrong.  x  tRipley's Believe It or Not, 4th Series, 1957. P. 15 asserts that the Count de Bouteville directed that his widow, age 20, should receive one gold piece during the first year of widowhood, the amount to be doubled each successive year she remained unmarried. She survived 69 years without marrying! Ripley says the Count 'never suspected the cumulative powers of arithmetical [sic!] progression'. 7.L.2.a.CHESSBOARD PROBLEM See Tropfke 630. See 5.F.1 for more details of books on the history of chess. Jens HQyrup. Subscientific mathematics: Undercurrents and missing links in the mathematical technology of the Hellenistic and Roman world. Preprint from Roskilde University, Institute of Communication Research, Educational Research and Theory of Science, 1990, Nr. 3. (Written for: Aufsteig und Niedergang der r?mischen Welt, vol. II 37,3 [??].) He discusses this type of problem, citing alUqldis [AbE alHassan Ahmad Ibn Ibrh3m alUql3dis3. Kitb al FusEl f3 al-Hisb al-Hind3. 952/953. MS 802, Yeni Cami, Istanbul. Translated and annotated by A. S. Saidan as: The Arithmetic of Al-Uqldis; Reidel, 1978. ??NYS. P. 337] as saying: "this is a question many people ask. Some ask about doubling one 30 times, and others ask about doubling it 64 times". HQyrup says that doubling 30 times is found in Babylonia, Roman Egypt, Carolingian France, medieval Damascus and medieval India. On pp. 2324, he describes the first two examples mentioned above and then mentions Alcuin and alUqldis. The last example is probably Bhaskara II. A cuneiform tablet from Old Babylonian Mari [Denis Soubeyran; Textes math)matiques de Mari; Revue d'Assyriologie 78 (1984) 1948. ??NYS. P. 30] has, in HQyrup's translation: "To one grain, one grain has been added: Two grains on the first day; Four grains on the second day; ...." this goes on to 30 days. The larger amounts are not computed as numbers, but converted to larger units. Old Babylonian is c1700. Papyrus Ifao 88 [B. Boyaval; Le P. Ifao 88: Probl/mes de conversion mon)taire; Zeitschrift fGr Papyrologie und Epigraphik 7 (1971) 165168, Tafel VI, ??NYS] starts with 5 and doubles 30 times, again using larger units for the later stages. HQyrup says this is a GrecoEgyptian papyrus 'probably to be dated to the Principate but perhaps as late as the fourth century' I am unable to determine what the Principate was. Perelman. FFF. 1934. 1957: prob. 52, pp. 7480; 1979: prob. 55, pp. 9298. = MCBF, prob.55, pp. 9098. This describes a Roman version where the general Terentius can take 1 coin the first day, 2 the second day, 4 the third day, ..., until he can't carry any more, which occurs on the 18th day. A footnote says this is a translation "from a Latin manuscript in the keeping of a private library in England." ?? Murray mentions the problem on pp. 51-52, 155, 167, 182 and discusses it in detail in his Chapter XII: The Invention of Chess in Muslim Legend, pp. 207-219. He discusses various versions of the invention of chess, some of which include the doubling reward. He describes the doubling legends in the following. al-YaqEb3 (c875). alMasudi (943). Firdaws3's Shhnma (1011). Kitb ash-shatranj [= AH] (1141) as AH f. 3b (= AbE Zakar3y [= H] f.6a). BM MS Arab. Add. 7515 (Rich) [= BM] (c1200?). von Eschenbach (c1220). BM Cotton Lib. MS Cleopatra, B.ix [= Cott.] (13C). ibn Khallikan (1256). Dante (1321). Shihbadd3n at-Tilimsni [= Man.] (c1370), which gives five versions. Kaj3n [= Y] (16C?). Murray 218 mentions two treatises on the problem: Al-Missis3. Tad3f buyEt ash-shatranj. 9C or 10C. Al-Akfn3. Tad3f adad ruqa ash-shatranj. c1340. On p. 217, Murray gives 10 variant spellings of Sissa and feels that Bland's connection of the name with Xerxes is right. On p. 218, he says the reward of corn would cover England to a depth of 38.4 feet. Murray 218. "This calculation is undoubtedly of Indian origin, the early Indian mathematicians being notoriously given to long-winded calculations of the character." He suggests the problem may be older than chess itself. Al-YaqEb3. Tar3kh. c875. Ed. by Houtsma, Leyden, 1883, i, 99-105. ??NYS. Cited by Murray 208 & 212. "Give me a gift in grains of corn upon the squares of the chessboard. On the first square one grain (on the second two), on the third square double of that on the second, and continue in the same way until the last square." [Quoted from Murray 213.] alMasudi (= Mas'udi = Ma'oudi). MurEj adh-dhahab [Meadows of gold]. 943. Translated by: C. Barbier de Meynard & P. de Courteille as: Les Prairies d'Or; Imprimerie Imp)riale, Paris, 1861. Vol. 1, Chap. VII, pp. 159-161. "The Indians ascribe a mysterious interpretation to the doubling of the squares of the chessboard; they establish a connexion between the First Cause which soars above the spheres and on which everything depends, and the sum of the square of its squares. This number equals 18,446,744,073,709,551,615 ...." [Quoted from Murray 210. The French ed. has two typographical errors in the number.] No mention of the Sessa legend. Muhammad ibn Ahmed AbElR3hn elB3rEn3 (= al-B3rEn3 = alBiruni). Kitb al-tr al-bq3ya an al-qurEn al-hal3ya (= al-thr al-bqiya min al-qurEn al khliya =Athr-ul-bkiya) (The Chronology of Ancient Nations). 1000. Arabic (and/or German??) ed. by E.Sachau, Leipzig, 1876 (or 1878??), pp. 138-139. ??NYS. English translation by E.Sachau, William H. Allen & Co., London, 1879, pp. 134-136. An earlier version is: E. Sachau; Algebraisches Gber das Schach bei B3rEn3; Zeitschr. Deut. Morgenlndischen Ges. 29 (1876) 148-156, esp. 151-155. Wieber, pp. 113-115, gives another version of the same text. Computes 1 + 2 + 4 + ... + 263 as 264 - 1 by repeated squaring. Doesn't mention Sessa. He shows the total is 2,305 mountains. "But these are (numerical) notions that the earth does not contain." Murray218 gives: "which is more than the world contains." but I'm not sure if alBiruni means the mountains or the numbers are more than earth can contain. BM MS Arab. Add. 7515 (Rich). Arabic MS with the spurious title "Kitb ash-Shatranj al Basr3", perhaps c1200. Copied in 1257. Described by Bland and Forbes, loc. cit. in 5.F.1 under Persian MS 211, and by Murray on p. 173. Murray denotes it BM. Bland, p. 26, says p. 6 of the MS gives the story of SCsah ben Dhir and the reward. Bland, p. 62, says the various forms of the name Sissah are corruptions of Xerxes. Forbes, pp. 74-76, does not mention the story or the reward. Murray 217 says all the Arabic MSS include the reward problem as part of one of their stories of the invention of chess, but on pp. 173 & 211-219, he doesn't mention the story in this MS specifically. However, on p. 173, he notes that the spurious first page gives the calculation in 15C Arabic and again in Turkish. Fibonacci. 1202. Pp. 309-310 (S: 435437). He induces the repeated squaring process and gets 264 - 1. He computes the equivalent number of shiploads of grain there is a typographic error in his result. Wolfram von Eschenbach. Willehalm. c1220. Ed. by Lachmann, p. 151, ??NYS quoted by Murray 755. "Ir hers mich bevilte, der Zende Ez zwispilte ame schchzabel iesl3ch velt mit cardam=m." Murray 755 gives several other medieval European references. (AlKdi Shemsedd3n Ahmed) Ibn Khallikan. Entry for: AbE Bakr as-SEli. In: Kitab wafayt al-ayn. 1256. Translated by MacGuckin de Slane as: Biographical Dictionary; (London, 1843-1871;) corrected reprint, Paris, 1868. Vol. III, p. 69-73. Sissah ibn Dhir, King Shihrm and the chessboard on pp. 69-71. An interpolation(?) mentions King Balhait. BM Cotton Lib. MS Cleopatra, B.ix. c1275. Anglo-French MS of c1275, described by Murray 579-580, where it is denoted Cott. No. 18, f.10a, gives doubling. Dante. Divina Commedia: Paradiso XXVIII.92. 1321. "Ed eran tante che'l numero loro Piu che'l doppiar degli scacchi s'imila." [Quoted from Murray 755.] Paolo dell'Abbaco. Trattato di Tutta l'Arta dell'Abacho. 1339. Op. cit. in 7.E. B 2433 f. 21v has an 8 x 8 board with two columns filled in with powers of two, starting with 2. Below he gives 264 and treats it as farthings and converts to danari, soldi, libri??, soldi d'oro, libri?? d'oro and then a further step that I cannot understand. No mention of chess or a reward. Thomas Hyde. Mandragorias seu Historia Shahiludii, .... (= Vol. 1 of De Ludis Orientalibus, see 4.B.5 for vol. 2.) From the Sheldonian Theatre (i.e. OUP), Oxford, 1694. Prolegomena curiosa. The initial material and the Prolegomena are unpaged but the folios of the Prolegomena are marked (a), (a 1), .... The material is on (d 1).r (d 4).v, which are pages 2532 if one starts counting from the beginning of the Prolegomena. He mentions Wallis (see below) and arithmetic (sic!) progressions and says the story is given in alSafadi (Salhadd3n as-Safad3 = alSphadi = AlSphadi) (d. 1363) in his Lmiyato l Agjam (printed various ways in the text). This must be his Sharh Lm3yat al-Ajam of c1350. Hyde gives some Arabic text and a Latin translation. Wallis gives the full Arabic text and translation. This refers to Ibn Khallikan. In his calculation, he uses various measures until he takes 239 grains as a granary, then 1024 granaries (= 249 grains) as a city, so the amount on the 64th square is 16384 (= 214) cities, but you know there are not so many cities in the whole world". He then gives 264 1 correctly and converts it into cubic miles, but seems off by a factor of ten see Wallis, below, who gives details of the units and calculations involved, noting that al-Safadi is finding the edge (= height) of a square pyramid of the volume of the pile of wheat. Hyde then adds a fragment from a Persian MS, Mugjizt, which gives the story with drachmas instead of grains of wheat, but the calculations are partly illegible. In his main text, pp. 3152 are on the invention of the game and he gives various stories, but doesn't mention the reward. Folkerts. Aufgabensammlungen. 1315C. 7 sources. Shihbadd3n AbEl-Abbs Ahmad ibn Yahya ibn Ab3 Hajala at-Tilimsni alH-anbal3. Kitb anmEdhaj al-qitl fi lab ash-shatranj (Book of the examples of warfare in the game of chess). c1370. Copied by Muhammed ibn Ali ibn Muhammed al-Arzag3 in 1446. This is the second of Dr. Lee's MSS, described in 5.F.1, denoted Man. by Murray. Described by Bland and Forbes, loc. cit. in 5.F.1 under Persian MS 211, and by Murray 175177 (as Man) & 207-219. Gives five versions of the chessboard story. The first is that of ibn Khallikan; others come from arRghib's Muhdart alUdab; from Qutbadd3n Muhammad ibn Abdalqdir's Durrat alMud3a and from alAkfn3. One calculates in lunar years and another version calculates in miles!! Columbia Algorism. c1350. Prob. 88, pp. 106-107. Chessboard. Uses repeated squaring. Copying error in the final value. Persian MS 211. Op. cit. in 5.F.1. c1400. Bland, loc. cit., p. 14, mentions "the well known story of the reward asked in grain". Forbes' pp. 64-66 is a translation of the episode of the Indian King Kaid's reward to Sassa. On p. 65, Forbes mentions various interpretations of the total. AR. c1450. Prob. 319, pp. 141, 180, 227. Chessboard, with very vague story. Benedetto da Firenze. Trattato di Praticha d'Arismetrica. Italian MS, c1464, Plimpton 189, Columbia University, New York. ??NYS. Chessboard. (Rara, 464-465; Van Egmond's Catalog 257258.) Pacioli. Summa. 1494. Ff. 43r43v, prob. 28. First mentions 1, 2, 6, 18, 54, ..., where each cell has double the previous total. Then does usual chessboard problem, but with no story. Computes by repeated squaring. Converts to castles of grain. Shows how to do 1 + 2 + 6 + 18 + ... for 64 cells and computes the result. Muhammad ibn Omar Kaj3n. Kitab al-munjih f3ilm ash-shatranj (A book to lead to success in the knowledge of chess). 16C? Translated into Persian by Muhammad ibn Husm ad-Daula, copied in 1612. Described by Bland and Forbes and more correctly by Murray on p. 179, where it is identified as MS BM Add. 16856 and denoted Y, since it was a present from Col. Wm. Yule. Bland, p. 20, mentions S1sah ben Dhir al Hindi and the reward claimed in grain. "The geometrical progression of the sixty four squares ... is computed here at full length, commencing with a Dirhem on the first square, and amounting to two thousand four hundred times the size of the whole globe in gold." Forbes describes this on pp. 76-77 and in the note on p. 65, where he computed the reward to make a cube of gold about 6 miles along an edge. He says the above Persian value is wrong somewhere, but he hasn't been able to see the original. [I can't tell if he means the Persian or the Arabic MS. If a dirhem was the size of an English 2p coin or an American quarter, the reward is about 2 x 104 km3, compared to earth's volume of about 1012 km3. The reward would make a cube about 27 km on an edge or about 17 miles on an edge.] Murray doesn't refer to this MS specifically. Ian Trenchant. L'Arithmetique. Lyons, 1566, 1571, 1578, ... ??NYS. 1578 ed., p. 297. 1,3,9,27. (H&S 91 gives French and English and says similar appear in Vander Hoecke (1537), Gemma Frisius (1540) and Buteo (1556).) Clavius. ??NYS. Computes number of shiploads of wheat required. (H&S 56.) van Etten. 1624. Prob. 87, pp. 111-118 (not in English editions). Includes chessboard as part XI, on p. 117. Henrion's Notte, p. 38, observes that there are many arithmetical errors in prob. 87 which the reader can easily correct. John Wallis. (Mathesis Universalis. T. Robinson, Oxford, 1657. Chap. 31.) = Operum Mathematicoroum. T. Robinson, Oxford, 1657. Part 1, chap. 31: De progressione geometrica, pp. 266285. This includes the story of Sessa and the Chessboard in Arabic & Latin, taken from alSafadi, c1350, giving much more text than Hyde (see above) does and explaining the units and the calculation, showing that alSafadi's 60 miles should be about 6 miles and this is the edge and height of a square pyramid of the same volume as the wheat. He then computes all the powers of two up to the 63rd and adds them! John Ayrton Paris [Philosophy in Sport made Science in Earnest; (Longman, Rees, Orme, Brown, and Green; London, 1827);, 8th ed., Murray, 1857, p.515] says Wallis got 9 English miles for the height and edge. Anonymous proposer; a Lady, solver, with Additional Solution. Ladies' Diary, 170910 = T. Leybourn, I: 34, quest. 6. 64 diamonds sold for 1 + 2 + 4 + ... + 263 grains of wheat. Suppose a pint of wheat contains 10,000 grains, a bushel of wheat weighs half a hundredweight [a hundredweight is 112 lb], the value is 5s per bushel, a horse can carry 1000 lb and a ship can carry 100 tons, then how much is the payment worth and how many horses or ships would be needed to carry it? Euler. Algebra. 1770. I.III.XI: Questions for practice, no. 3, p.170. Payment to Sessa, converted to value. Ozanam-Montucla. 1778. Prob. 3, 1778: 7678; 1803: 7881; 1814: 7072; 1840: 37-38. Problem wants the results of doublings, with no story. Discussion gives the story of Sessa, taken from Al-Sephadi. Gives various descriptions of the pile of grain, citing Wallis for one of these and says it would cover three times the area of France to a depth of one foot. Eadon. Repository. 1794. Pp. 369370, no. 11. Indian merchant selling 64 diamonds to a Persian king for grains of wheat, in verse. Supposing a pint holds 10000 grains and a bushel of 64 pints weighs 50 pounds, how many horse loads (of a thousand pounds each) does this make? How many ships of 100 tons capacity? Manuel des Sorciers. 1825. P. 84. ??NX Chessboard. The Boy's Own Book. The sovereign and the sage. 1828: 182; 1828-2:238; 1829 (US): 106; 1855: 393; 1868: 431. Uses 63 doublings for no reason. HuttonRutherford. A Course of Mathematics. 1841? Prob. 26, 1857: 82. Reward to Sessa for inventing chess. Takes a pint as 7680 grains and 512 pints as worth 27/6 to value the reward at 6.45 x 1012 . Walter Taylor. The Indian Juvenile Arithmetic .... Op. cit. in 5.B. 1849. P. 200. King conferring reward on a general. Computes number of seers, which contain 15,360 grains, and the value if 30 seers are worth one rupee. Nuts to Crack XIV (1845), no. 73. The sovereign and the sage. Almost identical to Boy's Own Book. Magician's Own Book. 1857. The sovereign and the sage, pp. 242243. A simplified version of OzanamMontucla. = Book of 500 Puzzles, 1859, pp. 5657. = Boy's Own Conjuring Book, 1860, p. 213. Vinot. 1860. Art. XVIII: Probl/me des )checs, pp. 3637. Uses grains of wheat and says there are 20,000 grains in a litre. Says the reward would cover France to a depth of 1.6 m. He gives the area of France as 9,223,372 km2. [Chambers]. Arithmetic. Op. cit. in 7.H. 1866? P. 267, quest. 54. Story of Sessa with grains of wheat. Suppose 7680 grains make a pint and a quarter is worth 1 7s 6d, how much was the wheat worth? James Cornwell & Joshua G. Fitch. The Science of Arithmetic: .... 11th ed., Simpkin, Marshall, & Co., London, et al., 1867. (The 1888 ed. is almost identical to this, so I suspect they are close to identical to the 2nd ed. of 1856.) Exercises CXLIII, no. 6, pp. 299 & 371. Chessboard problem with no story, assumes 7680 grains to a pint. Mittenzwey. 1880. Prob. 94, pp. 19 & 68; 1895?: 109, pp. 24 & 71; 1917: 109, pp. 22 & 68. King Shehran rewarding Sessa Eba Daher, according to Asephad. Cassell's. 1881. P. 101: Sovereign and the sage. Uses sage's and king's common age of 64, with no reference to chessboard. Lucas. L'Arithm)tique Amusante. 1895. Le grains du bl) de Sessa, pp. 150151. Says it would take 8 times the surface of the earth to grow enough grain. Berkeley & Rowland. Card Tricks and Puzzles. 1892. The chess inventor's reward, pp. 112-114. Assumes 7489 29/35 grains to the pound, with 112 pounds to the cwt., 20 cwt. to the ton and 1024 tons to the cargo, getting 1,073,741,824 cargoes, less one grain. The number of grains is chosen so that a ton contains exactly 224 grains of rice and the answer is 230 cargoes less one grain. 7.L.2.b.HORSESHOE NAILS PROBLEM See Tropfke 632. AR. c1450. Prob. 274, 317, 318, 353. Pp. 125, 140, 154, 180, 227.  x  th274: 32 horseshoe nails. 317: 16 cow nails. 318: 32 horseshoe nails. 353: 28 nails text is obscure.  x  tRiese. Rechenung nach der lenge .... 1525. (Loc. cit. under Riese, Die Coss.) Prob. 32, p. 20. 32 horseshoe nails. Christoff Rudolff. KGnstliche rechnung mit der ziffern und mit den zal pfenninge. Vienna, 1526; NGrnberg, 1532, 1534, et seq. F.N.viii.v. ??NYS. 32 horseshoe nails. (H&S56 gives German.) Apianus. Kauffmanss Rechnung. 1527. Ff. D.vi.r D.vi.v. 32 horseshoe nails. Anon. Trattato d'Aritmetica, e del Misure. MS, c1535, in Plimpton Collection, Columbia Univ. ??NYS. Horseshoe problem: 1 + 2 + 4 + ... + 223. (Rara, 482-484, with reproduction on p.484.) Recorde. First Part. 1543. Ff. L.ii.r L.ii.v (1668: 141142: A question of an Horse). 24horseshoe nails. Buteo. Logistica. 1559. Prob. 34, pp. 237238. 24 horseshoe nails. (H&S 56.) van Etten. 1624. Prob. 87, pp. 111-118 (not in English editions). Includes 24 horseshoe nails problem as part VII on p. 115. Henrion's Notte, p. 38, observes that there are many arithmetical errors in prob. 87 which the reader can easily correct. Wells. 1698. No. 102, p. 205. 24 nails. Ozanam. 1725. Prob. 11, question 4, 1725: 77-78 & 80. Part of prob. 3, 1778: 7980; 1803:81; 1814: 72; 1840: 38. 24 nails first asks for the price of the 24th, then the total. Dilworth. Schoolmaster's Assistant. 1743. P. 96, no. 1. 32 nails, starting with a farthing. Walkingame. Tutor's Assistant. 1751. Geometrical Progression, prob. 6, 1777: p. 95; 1835: p. 103; 1860: p. 123. 32 nails, one farthing for the first, wants total, which he gives in s/d. Mair. 1765? P. 493, ex. III. "What will a horse cost by tripling the 32 nails in his shoes with a farthing?" I.e., 32 horseshoe nails, but with tripling! Euler. Algebra. 1770. I.III.XI.511, p. 166. Horse to be sold for the value of 32 nails, 1penny for the first, .... Vyse. Tutor's Guide. 1771? Prob. 2, 1793: p. 140; 1799: p. 148 & Key p. 190. 36 horseshoe nails. Want value of last one, starting , , 1, .... Bullen. Op. cit. in 7.G.1. 1789. Chap. 31, prob. 1, p. 215. Same as Walkingame. Bonnycastle. Algebra. 10th ed., 1815. P. 80, no. 6. Same as Walkingame. Manuel des Sorciers. 1825. P. 84. ??NX 24 horseshoe nails. The Boy's Own Book. The horsedealer's bargain. 1828: 182; 18282:238; 1829 (US): 106; 1843 (Paris): 346; 1855: 393-394; 1868: 431432. Wants value of 24th nail, starting with a farthing. =Boy's Treasury, 1844, p. 304. = de Savigny, 1846, p. 292: Le march) aux chevaux. Nuts to Crack XIV (1845), no. 74. The horsedealer's bargain. Almost identical to Boy's Own Book. Walter Taylor. The Indian Juvenile Arithmetic .... Op. cit. in 5.B. 1849. P. 199. 32 nails. [Chambers]. Arithmetic. Op. cit. in 7.H. 1866? P. 268, quest. 64. 24 nails, starting with a farthing. Finds total. Mittenzwey. 1880. Prob. 93, pp. 19 & 68; 1895?: 108, pp. 2324 & 71; 1917: 109, pp. 2122 & 68. 32 horseshoe nails, starting at 1 pf. Cassell's. 1881. P. 101: The horse-dealer's bargain. 24 nails, unclear, but uses 223 farthings as the answer. 7.L.2.c. USE OF 1, 2, 4, ... AS WEIGHTS, ETC.  tSee Tropfke 633. A special case of this is the use of such amounts to make regular unit payments, e.g. rent of one per day. See: Knobloch; Fibonacci; BR; Widman; Tartaglia; Gori; Les Amusemens.  t Eberhard Knobloch. Zur Fberlieferungsgeschichte des Bachetschen Gewichtsproblems. Sudhoffs Archiv 57 (1973) 142151. This describes the history of this topic and 7.L.3 from Fibonacci to Ozanam (1694). He gives a table showing occurrences of: powers of two, powers of three, weight problem, payment problem. I am not entirely clear what he means in the first three cases I would have two kinds of weight problem corresponding to the first two cases and perhaps some of his references in the first case are listed under 7.L.2. However, the last case clearly corresponds to the problem of making a payment of one unit per day as in Fibonacci. He lists this as occurring in Fibonacci, BR, Widmann and Tartaglia and notes that Sanford, H&S 91, only noticed Fibonacci. Knobloch notes that Ball's citations are not very good and that Ahrens' note about them does not go much deeper. I have a number of references listed below which were not available to Knobloch. Fibonacci. 1202. P. 298 (S: 421). Uses 5 ciphi of value 1, 2, 4, 8, 15 to pay a man at rate of 1 per day for 30 days. BR. c1305. No. 93, pp. 112-113. Use of 1, 2, 4 as payments at rate of one per year for 7 years. Widman. Op. cit. in 7.G.1. 1489. Ff. 138v139r. ??NYS Knobloch says he uses values of 1, 2, 4, 8, 16 to pay for 31 days. Pacioli. Summa. 1494. Ff. 97v98r, no. 35. Use five cups to pay daily rent for 30 days. Uses cups of weight 1, 2, 4, 8, 15. In De Viribus, c1500, F. XIIIv, item 86 in the Indice for the third part is: De 5 tazze, diversi pesi ogni di paga l'oste (Of 5 cups of diverse weights to pay the landlord every day) = Peirani 20, but at the end Pacioli says this problem is in 'libro nostro', i.e. the Summa. Cf Agostini, p. 6. Tartaglia. General Trattato, 1556, part 2, book 1, chap. 16, art. 32: Di una particolar proprieta della progression doppia geometrica, p. 17v. Weights: 1, 2, 4, 8, ... (See MUS I 89.). Also does payments with 1, 2, 4, 8, 16, 29. Knobloch also refers to art. 3335 ??NYS and notes that the folios are misnumbered, but miscites 'doppia' as 'treppia' here. This covers the powers of 3 also. Buteo. Logistica. 1559. Prob. 91, pp. 309312. Use of 1, 2, 4, 8, 16, ... as weights. (Cited by Knobloch.) Knobloch also cites Ian Trenchant (1566), Daniel Schwenter (1636), Franz van Schooten (1657). Gori. Libro di arimetricha. 1571. Ff. 71r-71v (p. 76). Use of cups weighing 1, 2, 4 to make all weights through 7, to pay for days at one per day. Bachet. Problemes. 1612. Addl. prob. V & V(bis), 1612: 143146; as one prob. V, 1624:215219; 1884: 154156. Mentions 1,2,4,8,16 and cites Tartaglia, art. 32 only. This was omitted in the 1874 ed. Knobloch cites 1612, pp. 127 & 143146, but but p. 127 is Addl. prob. I, which is a Chinese Remainder problem? van Etten/Henrion. 1630. Notte to prob. 53, pp. 20-21. Refers to Bachet and compares with ternary weights. Ozanam. 1694.  x  thProb. 8, 1696: 3335; 1708: 2932. Prob. 11, 1725: 6875. Section II, 1778: 6874; 1803: 7076; 1814: ??NYS; 1840: 3436. A discussion of geometric progression and a mention of 1,2,4, ..., without any application to weighing. 1778 et seq. also mentions 1, 3, 9, .... Prob. 12, vol. II, 1694: 1819 (??NYS). Prob. 12, 1696: 284 & fig. 131, plate 46, p.275; 1708: 360 & fig. 26, plate 14, opp. p. 351. Prob. 8, vol.II, 1725: 345-348 & fig. 131, plate 46 (42). Prob. 14, vol. I, 1778: 206-207; 1803: 201202. Prob. 13, vol. II, 1814: 174175; 1840: 90-91. Gives double and triple progressions. Knobloch gives the 1694 citation. The figure is just a picture of a balance and is not informative the same figure is also cited for other sets of weights.  x  tLes Amusemens. 1749. Prob. 8, p. 128. Coins of value 1, 2, 4, 8, 15 to pay for a room at a rate of 1 per day for 30 days. The Bile Beans Puzzle Book. 1933. No. 42: Money juggling. Place 1000 in 10 bags so any amount can be paid without opening a bag. Solution has bags of: 1, 2, 4, 8, 16, 32, 63, 127, 254, 493. I cannot see why the solution isn't: 1, 2, 4, 8, 16, 32, 64, 128, 256, 489. 7.L.3.1 + 3 + 9 + ... AND OTHER SYSTEMS OF WEIGHTS See MUS I 8898; Tropfke 633. Tabari. Mift]h almu]mal]t. c1075. P. 125ff., no. 43. ??NYS Hermelink, op. cit. in 3.A, and Tropfke 634635 say this gives 1, 3, 9, ..., 19683 = 39 to weigh up to 10,000. Fibonacci. 1202.  x  thP. 297 (S: 420421). Weights 1, 3, 9, 27 and 1, 3, 9, 27, 81 'et sic eodem ordine possunt addi pesones in infinitum' [and thus in the same order weights can be added without end]. Pp. 310-311 (S: 437). Finds 1 + 2 + 6 + 18 + ... + 2*362 = 363 by repeated squaring to get 364 and then divides by 3.  x  tGherardi. Libro di ragioni. 1328. P. 53. Weights 1, 3, 9, 27, 80 to weigh up through 120. Columbia Algorism. c1350. Prob. 71, pp. 92-93. Weights 1, 3, 9, 27. AR. c1450. Prob. 127, pp. 67 & 182. 1, 3, 9, 27. Chuquet. 1484. Prob. 142.  x  th1, 2, 7 to weigh up to 10 (English in FHM 225); 1, 2, 4, 15 up to 22; 1, 3, 9 up to 13; 1, 3, 9, 27 up to 40; (original of this and the next case reproduced on FHM 226) 1, 3, 9, 27, 81 up to 121. Knobloch says Chuquet gives a general solution, but I don't see that Chuquet is general.  x  tPacioli. Summa. 1494. Ff. 97r97v, no. 34. General discussion of 1, 3, 9, 27, 81, 243, .... In De Viribus, c1500, F. XIIIv, item 85 in the Indice for the third part is: De far 4 pesi che pesi fin 40 (To make four weights which weigh to 40) = Peirani 20, but at the end Pacioli says this problem is in 'libro nostro', i.e. the Summa. Cf Agostini, p. 6. Cardan. Practica Arithmetice. 1539. Chap. 65, section 12, ff. BB.vii.r BB.vii.v (p.136). Weights 1, 3, 9, 27, .... Knobloch also cites: Giel vanden Hoecke (1537); Gemma Frisius (1540); MichaelStifel(1553); Simon Jacob (1565); Ian Trenchant (1566); DanielSchwenter(1636); Kaspar Ens (1628); Claude Mydorge (1639); FransvanSchooten (1657). Tartaglia, 1556 see in 7.L.2.c. Buteo. Logistica. 1559. Prob. 91, pp. 309312. Use of 1, 3, 9, 27, ... as weights. (Cited by Knobloch.) John [Johann (or Hanss) Jacob] Wecker. Eighteen Books of the Secrets of Art & Nature Being the Summe and Substance of Naturall Philosophy, Methodically Digested .... (As: De Secretis Libri XVII; P. Perna, Basel, 1582 ??NYS) Now much Augmented and Inlarged by Dr. R. Read. Simon Miller, London, 1660, 1661 [Toole Stott 1195, 1196]; reproduced by Robert Stockwell, London, nd [c1988]. Book XVI Of the Secrets of Sciences: chap. 19 Of Geometricall Secrets: To poyse all things by four Weights, p.289. 1, 3, 9, 27; 1, 3, 9, 27, 81; 1, 3, 9, 27, 81, 243. Cites Gemma Frisius. Bachet. Problemes. 1612. Addl. prob. V & V(bis), 1612: 143146; as one prob. V, 1624:215219; 1884: 154-156. Weights: 1, 3, 9, 27, ..., and the general case via the sum of a GP. In the 1612 ed., Bachet only does the cases 40 and 121, then does the general case. Knobloch cites 1612, pp. 127 & 143146, but p. 127 is Addl. prob. I, which is a Chinese Remainder problem. He also says this is the first proof of the problem, excepting Chuquet, though I don't see such in Chuquet. van Etten. 1624. Prob. 53 (48), pp. 48-49 (72). 1, 3, 9, 27; 1, 3, 9, 27, 81; 1,3,9,27,81,243. Henrion's Notte, pp. 20-21, refers to Bachet and compares this with binary weights. Ozanam. 1694.  x  thProb. 8, 1696: 3335; 1708: 2932. Prob. 11, 1725: 6875. Section II, 1778: 6874; 1803: 7076; 1814: ??NYS; 1840: 3436. A discussion of geometric progression and a mention of 1,2,4, ..., without any application to weighing. 1778 et seq. also mentions 1, 3, 9, .... Prob. 12, vol. II, 1694: 1819 (??NYS). Prob. 12, 1696: 284 & fig. 131, plate 46, p.275; 1708: 360 & fig. 26, plate 14, opp. p. 351. Prob. 8, vol.II, 1725: 345-348 & fig. 131, plate 46 (42). Prob. 14, vol. I, 1778: 206-207; 1803: 201202. Prob. 13, vol. II, 1814: 174175; 1840: 90-91. Gives double and triple progressions. Knobloch gives the 1694 citation. The figure is just a picture of a balance and is not informative the same figure is also cited for other sets of weights.  x  tLes Amusemens. 1749. Prob. 18, p. 140: Les Poids. Weights 1, 3, 9, 27, 81, 243. Vyse. Tutor's Guide. 1771? Prob. 2, 1793: p. 303; 1799: p. 316 & Key pp. 356357. Weights 1, 3, 9, 27. Bonnycastle. Algebra. 1782. P. 202, no. 13. 1, 3, 9, 27, 81, 243, 729, 2187 to weigh to 29 hundred weight an English hundred weight is 112 pounds. c= 1815: p. 230, no. 33. 1, 3, 9, 27, 81 to weigh to a hundred weight. Eadon. Repository. 1794. Pp. 297298, no. 1. 1, 3, 9, ..., 313. Badcock. Philosophical Recreations, or, Winter Amusements. [1820]. P. 50, no. 77: To find the least Number of Weights that will weigh from One Pound to Forty. 1, 3, 9, 27. Jackson. Rational Amusement. 1821. Curious Arithmetical Questions. No. 24, pp. 20 & 78-79. 1, 2, 4, 8, 16, ... and 1, 3, 9, 27, 81, .... Rational Recreations. 1824. Exer. 22, p. 131. 1, 3, 9, 27. Endless Amusement II. 1826?  x  thP. 105: To find the least Number of Weights that will weigh from One Pound to Forty. = Badcock. Prob. 24, p. 201. To name five weights, which, added together, make 121 pounds; by means of which may be weighed any intermediate weight, excluding fractions. 1,3, 9, 27, 81. = New Sphinx, c1840, p. 137.  x  tYoung Man's Book. 1839. P. 242. To name five weights, .... Identical to Endless Amusement II, p. 201. Boy's Own Book. 1843 (Paris): 346347. "To find the least number of weights which will weigh any intermediate weight, from one pound to forty, exclusive of fractions. Indicates that one can continue the progression." = Boy's Treasury, 1844, p.304. Family Friend 1 (1849) 150 & 178. Problems, arithmetical puzzles, &c. 5. How can one divide 40 lb into four weights to weigh every value from 1 to 40? Proposer says he can't do it. Usual answer, but one solver gives weights 6, 10, 11, 13. However, the latter weights will not weigh 22, 25, 26, 31, 32, 33, 35, 36, 37, 38, 39. Boy's Own Book. To weigh from one to forty pounds with four weights. 1855: 392; 1868:430. 1, 3, 9, 27. No generalizations. Magician's Own Book. 1857. The mathematical blacksmith, p. 230. 1, 3, 9, 27 and this can be continued. = Boy's Own Conjuring Book, 1860, p. 200. Todhunter. Algebra, 5th ed. 1870. Miscellaneous Examples, no. 175, p. 559. Show that 1,1, 5, 5, 25, 25, 125, 125 can weigh any integral amount up to 312. No solution given. F. J. P. Riecke. Op. cit. in 4.A.1, vol. 3, 1873. Art. 3: Die Zauberkarten, p. 13. Uses balanced ternary for divination. See under 7.M.4 Mittenzwey. 1880.  x  thProb. 107, pp. 22 & 74; 1895?: 124, pp. 26 & 7677; 1917: 124, pp. 24 & 7475. Stone of weight 40 breaks into four parts which weigh up through 40. Solution is a table showing how to weigh up through 20. 1895? adds solutions for weights 1, 2, 3, 34; 5, 9, 10, 16; allowing several weighings, e.g. 4 is obtained by weighing out 2 twice. 1895?: prob. 125, pp. 26 & 77; 1917: 125, pp. 24 & 75. Five weights to get through 121.  x  tP. A. MacMahon. Certain special partitions of numbers. Quart. J. Math. 21 (1886) 367-373. Very technical. P. A. MacMahon. Weighing by a series of weights. Nature 43 (No.1101) (4 Dec 1890) 113-114. Less technical description of the above work. Lucas. L'Arithm)tique Amusante. 1895. Pp. 166168. Notes that pharmacists, etc. use weights: 1, 1, 2, 5, 10, 10, 20, 50, 100, 100, 200, 500, 1000, 1000, 2000, 5000, .... Discusses ternary. Wehman. New Book of 200 Puzzles. 1908. P. 49. 1, 3, 9, 27, 81. Ahrens. MUS I. 1910. Pp. 8898 discusses this and some generalizations like MacMahon's. In 2002, Miodrag Novakovia told me that a student had told him how to determine twice as many integral weights with the same number of weights. E.g., he used weights 2, 6, 18, 54 to determine integral weights 1, 2, ..., 81. One can get exact balancing for the even values: 2, 4, ..., 80. The odd values fall between two consecutive even amounts, so if a package weighs more than 6 but less than 8, we deduce it weighs 7! 7.M.BINARY SYSTEM AND BINARY RECREATIONS  tThe binary system has several origins. a)Egyptian & Russian peasant multiplication. b)Weighing see 7.L.2.c. c)Binary divination see 7.M.4. d)The works below. See also: 5.E.2 for Memory Wheels; 5.F.4 for circuits on the n-cube; 5.AA for an application to cardshuffling; 7.AA.1 for Negabinary.  t Anton Glaser. History of Binary and Other Nondecimal Numeration. Published by the author, 1971; (2nd ed., Tomash, Los Angeles, 1981). General survey, but has numerous omissions see the review by Knuth at Harriot, below, and MR 84f:01126. He has no references to early Chinese material. Shao Yung. c1060. Sung Yuan HsGeh An, chap. 10. Fu-Hsi diagram of the 64 hexagrams of the I-Ching, in binary order. A version appears in Leibniz-Briefe 105 (Bouvet) Bl. 27r/28r in the Niederschsische Landesbibliothek, Hannover. Needham, vol. 2, p. 341, notes that this had only been published in Japanese and Chinese up to 1956. See Zacher & Kinz= below for reproductions. Also reproduced in: E.J. Aiton; Essay Review [of Zacher, below]; Annals of Science 31 (1974) 575-578. Chu Hsi. Chou I Pen I Thu Shuo. 12C. Fu-Hsi Liu-shih-ssu Kua Tzhu HsG (Segregation Table of the symbols of the Book of Changes) reproduced in Hu Wei's I Thu Ming Pien. An illustration is given in Needham, vol. 2, fig. 41 = plate XVI, opp. p.276 he says it is based on the original chart of Shao Yung and that Tshai Chhen (c1210) gave a simplified version. Also in Kinz= and in Aiton & Shimao, below. Shows the alternation of 0 and 1 in each binary place. Thomas Harriot. Unpublished MS. c1604. Described by J. W. Shirley; Binary numeration before Leibniz; Amer. J. Physics 19 (1951) 452-454; and by D. E. Knuth; Review of 'History of Binary and Other Nondecimal Numeration'; HM 10 (1983)) 236-243. This shows some binary calculation. Shirley reproduces BM: Add MSS 6786, ff. 346v-347r. Knuth cites 6782, 1r, 247r; 6786, 243v, 305r, 346v, 347r, 516v; 6788, 244v. Francis Bacon. Of the Advancement of Learning. 1605. ??NYS. Describes his binary 5-bit coding. Francis Bacon. De augmentis scientarum. 1623. ??NYS. Full description of his coding. He does not have any arithmetic content, so he is not really part of the development of binary. John Napier. Rabdologiae. Edinburgh, 1617. ??NYS. Describes binary as far as extracting square roots. William F. Hawkins; The Mathematical Work of John Napier (1550-1617); Ph.D. thesis, Univ. of Auckland, 1982, ??NYS, asserts this is THE invention of the binary system. G. W. Leibniz. De Progressione Dyadica. 3pp. Latin MS of Mar 1679. Facsimile and translation into German included in: Herrn von Leibniz' Rechnung mit Null und Eins; Siemens Aktiengesellschaft, (1966), 2nd corrected printing, 1969; facsimile between pp.20 & 21 and translation on pp. 4247, with an essay by Hermann J. Greve: Entdeckung der binren Welt on pp. 2131. His first, unpublished, MS on the binary system, showing all the arithmetic processes. H. J. Zacher. Die Hauptschriften zur Dyadik von G. W. Leibniz. Klosterman, Frankfurt, 1973. Gathers almost all the Leibniz material, notably omitting the above 1679 paper. He does reproduce the Fu-Hsi diagram sent by Bouvet (cf. Shao Yung above). However Leibniz's letter of 2 Jan 1697 to Herzog Rudolf Augustus, in which he gives his drawing of his plan for a medallion commemorating the binary system, is now lost, but it was published in 1734. G. W. Leibniz. Two Latin letters on the binary system, 29 Mar 1698 & 17 May 1698, recipient not identified, apparently the author of a book in 1694 which occasioned Leibniz's correspondence with him. Opera Omnia, vol. 3, 1768, pp. 183190. Facsimile and translation into German included in: Herrn von Leibniz' Rechnung mit Null und Eins; Siemens Aktiengesellschaft, 2nd corrected printing, 1966; facsimile between pp.40 & 41 and translation on pp. 5360. On p. 26, it seems to say the second letter was sent to Johann Christian Schulenberg. G. W. Leibniz. Explication de l'arithm/tique binaire. Histoire de l'Academie Royale des Sciences 1703 (1705) 8589. Facsimile and translation into German included in: Herrn von Leibniz' Rechnung mit Null und Eins; Siemens Aktiengesellschaft, 2nd corrected printing, 1966; facsimile between pp. 32 & 33 and translation on pp. 4852. Illustrates all the arithmetic operations and discusses the Chinese trigrams of 'Fohy' and his correspondence with Father Bouvet in China. G. W. Leibniz. Letter of 1716 to Bouvet. ??NYS cited in Needham, vol. 2, p. 342. Fourth section is: Des Caract/res dont Fohi, Fondateur de l'Empire Chinois, s'est servi dans ses Ecrits, et de l'Arithm)tique Binaire. Gorai Kinz=. Juky= no Doitsu seiji shis= ni oyoboseru eiky= (Influence of Confucianism on German Political Thought [in Japanese]). Waseda Univ. Press, Tokyo, 1929. ??NYS. First publication of Leibniz's correspondence with Bouvet which led to the identification of the Fu Hsi diagram with the binary numbers. Gives a redrawn Fu-Hsi diagram and a segregation table. E. J. Aiton & Eikon Shimao. Gorai Kinz='s study of Leibniz and the I Ching hexagrams. Annals of Science 38 (1981) 71-92. Describes the above work. Reproduces Kinzo's Fu-Hsi diagram and segregation table. Ahrens. MUS I. 1910. 24104 discusses numeration systems in general and numerous properties of binary and powers of 2. Gardner. SA (Aug 1972) c= Knotted, chap. 2. General survey of binary recreations. The material in the book is much expanded from the SA column. 7.M.1.CHINESE RINGS See MUS I 6172; S&B 104-107, 111 & 135. See also 4.A.4, 11.K.1. In various places and languages, the following names are used: Chinese Rings Chainese Rings [from www.tama.or.jp/~tane, via Dic Sonneveld, 13 Nov 2002] Cardan's Rings, but Cardan called it Instrumentum ludicrum RyouKaikTjyo or Lau Kk Ch'a = Delayguestinstrument Kau Tsz' Lin Wain = Nine connected rings Chienowa = Wisdom rings Kyrenkan = Nine connected rings Lien nuan chhuan [from www.roma.unisa.edu.au, via Dic Sonneveld, 13 Nov 2002] Tarr