WPCL √ 2BJ|x╒!╨ x ╨╨╨№Ё ф ╪                                                                    ╨╨КВ╨╚╨┴`┴SOURCES й page !╒╨ ░x ╨╨╨░дШМ А t                                                                    ░╨╨КВ╨╚╨ ╨╨дШМ А t                                                                      ░╨╨├ ├7.┴┴ARITHMETIC & NUMBER-THEORETIC RECREATIONS─ ─ ┴┴├ ├7.A.┴┴FIBONACCI NUMBERS─ ─ ╨╨░дШМ А t                                                                    ░╨╨┴┴We use the standard form: F├├0── = 0, F├├1── = 1, F├├n+1── = F├├n── + F├├nй1──, with the auxiliary Lucas numbers being given by: L├├0── = 2, L├├1── = 1, L├├n+1── = L├├n── + L├├nй1──. ╨╨дШМ А t                                                                      ░╨╨ Parmanand Singh. The soйcalled Fibonacci numbers in ancient and medieval India. HM 12 (1985) 229й244. 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 F├├n+1──. ┴┴┴┴Pingala (cй450) 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 F├├n+1── = F├├n── + F├├nй1──. 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 kйth 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 F├├n+1── = └$└├├i── BC(nйi,i). ┴┴┴┴Narayana Pandita (= N└]└r└]└yana Pandita) 's Ganita Kaumudi (1356) studies additive sequences in chap.а13, where each term is the sum of the last q terms. He gives rules which are equivalent to finding the coefficients of (1 + x + ... + x├├qй1──)├├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 йй quoted by Kripa Shankar Shukla in the Introduction to his edition of: Narayana Pandita (= N└]└r└]└yana Pandita); B└Е└jaganit└]└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. 160й172. 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 HerzйFischler. De quand date le premier rapprochement entre la suite de Fibonacci et la division en extr└+└me et moyenne raison? Centaurus 28 (1985) 129й138. Discusses the history of the result that the ratio F├├n+1──/F├├n── 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 HerzйFischler. 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. 157й162 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) 2й19. This surveys many of the common misconceptions йй 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 йй with 59 references. He also discusses the origin of the term 'golden section', sketching the results given in HerzйFischler's book. Thomas Koshy. Fibonacci and Lucas Numbers with Applications. WileyйInterscience, 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 secondйhand 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: 404й405): 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 йй the pair propagate in the first month so there are F├├n+2── pairs at the end of the nйth month. (English translation in: Struik, Source Book, pp.а2-3.) I have colour slides of this from L.IV.20 & 21 and Conventi Soppresi, C. I. 2616. This is on ff.а130rй130v 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 & HerzйFischler and discussed in HerzйFischler's book, pp. 157й158. II.11 involves division in mean and extreme ratio. Uses 89, 144, 233 and that 144├├2── = 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 M└└stlin. ??NYS йй described in HerzйFischler's book, p. 158. This gives a construction for division in extreme and mean ration. On the original, M└└stlin has added his numerical calculations, getting 1/└C└ = .6180340, which HerzкFischler believes to be the first time anyone actually calculated this number. J. Kepler. Letter of 12 May 1608 to Joachim Tanckius. ??NYS йй described in HerzйFischler (1986), Curchin & HerzйFischler and HerzйFishler's book, pp. 160й161. Shows that he knows that the ratio F├├n+1──/F├├n── approaches └C└ and that F├├n──├├2── + (й1)├├n── = F├├nй1──F├├n+1──. J. Kepler. The Six-Cornered Snowflake. Op. cit. in 6.AT.3. 1611. P.а12 (20-21). Mentions golden section in polyhedra and that the ratio F├├n+1──/F├├n── approaches └C└. See HerzкFischler's book, p. 161. Albert Girard, ed. Les └ж└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 F├├n+1──/F├├n── 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 йй but he says 13,а13, 21 'rather precisely constitutes an isosceles triangle having the angle of a pentagon'. HerzйFischler'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.а128-134. Describes how to find the generating function of a recurrence. One of his illustrations is the Lucas numbers for which he gets: ┴┴┴┴x + 3x├├2── + 4x├├3── + 7x├├4── + 11x├├5── + ... = (2x + x├├2──)/(1 й x й x├├2──). 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) a├├n── 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, Birkh└└user, 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 F├├n──. [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. 193й206. 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 Am├├n── + Bp├├n── 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 F├├n──├├2──а+а(-1)├├n──аа=ааF├├nй1──F├├n+1──. This says that the triple F├├nй1──, F├├n──, F├├n+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 F├├n+1──/F├├n── does converge to └C└. HerzйFischler'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 F├├n──├├2──а+а(-1)├├n──аа=ааF├├nй1──F├├n+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, 1768й69 = 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 + ... + F├├39──. We would give this as: F├├41── й 1, but he gets it as: 2F├├39── + F├├38── й 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 ElementarйMathematik. 2nd ed., Jonas VerlagsйBuchhandlung, Berlin, 1835. P. 194, footnote to Prop. 5. ??NYS йй extensively discussed in HerzкFischler'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. HerzйFischler 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 & H└└rtel, Leipzig, 1876. ??NYS. Origin of the aesthetic experiments on golden rectangles. Koshy [p. 5] says Lucas originally called F├├n── 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) 70й75; which might be the desired paper. ┴┴┴┴L'arithm└)└tique, la s└)└rie de Lam└)└, le probl└/└me de BehaйEddin, etc. Nouvelles Annales de Math└)└matiques 15 (1876) 20pp. └(└douard Lucas. Th└)└orie des fonctions num└)└riques simplement p└)└riodiques. AJM 1 (1878) 184й240 (Sections 1й23) & 289й321 (Sections 24й30). [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 1877й1878 with some corrections and additions. Robert D. Carmichael; Annals of Math. (2) 15 (1913) 30й70, ??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 nйth term is F├├nй2──A + F├├nй1──B and the sum of the first 2n terms is F├├2n──A + (F├├2n+1──й1)B = L├├n── (F├├n──A + F├├n+1──B), 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─ ─ ╨╨░дШМ А t                                                                    ░╨╨┴┴See 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 йй 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└е└, Cardan, Coburg, Bachet, van Etten, Yoshida, Muramatsu, Schnippel, Ozanam (1696 & 1725), LesаAmusemens, Fujita, Euler, Miyake, Matuoka, Boy's Own Book, Nuts to Crack, TheаSociable, 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, ╨╨дШМ А 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.аD. 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, ╨╨░дШМ А t                                                                    ░╨╨ ┴┴There are a few examples where one counts down to the last two persons йй 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. ╨╨дШМ А 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, Codexаlat.аMonacensisа14908, Chuquet, Tartaglia, Buteo, Hunt, Yoshida, Muramatsu, Wingate/Kersey, Schnippel, Alberti, Shinpen Kinkoйki, Fujita, Miyake, Matuoka, SanpoаChieаBukuro, 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 йй ??check: Codex lat. Monacensis 14836, Meermanische Codex, at-Tilims└└ni, 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, n├├2──йn+1 & nй1 counted by n: Lucas (1894), Cesar└A└, Franel, Akar, ╨╨░дШМ А 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,аe, 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,а5, 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 ?? йй 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 йй to be added below. Hyde gives an Arabic mnemonic due to alйSafadi using the first letters of the Arabic alphabet: a, b, gj, d, h. ╨╨дШМ А t                                                                      ░╨╨Dahbagja Ababgja Baba [= Da h b a gj a A ba b gj a Ba b a]: Hyde from alйSafadi. 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; Ozanamа1725; LesаAmusemens; Hooper; Jackson; Manuel des Sorciers; Boy's Own Book; TheаSociable; 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; LesаAmusemens; 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.а14. 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. 118й169. 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 йй 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 AngloйSaxon for 'to play']) = Sankt Peter Lek or Sankt P└└ders Lek (in Swedish). Murphy cites Schnippel & Bolte to assert that it was known to the Arabs in the 14C йй 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 Dubh└└in Anso (Here is the Thieving of Fionn and Dubh└└n). 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 B└└n (Ceann Dubhrann of the White Sandhills). The Gaelic names Fionn and Dubh└└n 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.а1409 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 f└G└r Philologie (NS) 9 (1854) 296-301, with material of interest on pp. 298й299. Latin given in: M.аCurtze, 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 M└G└nchen; AGM 7 (1895) 105 & 111-112 (Supplement to Zeitsch. f└G└r 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.аcit. 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. ╨ дx ╨╨╨ШМ А th                                                                      д╨╨Meermanische Codex, 10C. Mentions counting by other values. Leiden Miscellancodex, 12C Basel Miscellancodex, 13C ╨ ░x ╨╨╨дШМ А t                                                                      ░╨╨Michinori Fujiwara (1106й1159). This work is lost, but has been conjectured to contain a form of the problem йй see under Kenk└е└, c1331, and Yoshida, 1634. Rabbi Abraham ben Ezra. Ta'hbula (or Tachb└E└la), c1150. ??NYS йй described in: Moritz Steinschneider; Abraham ibn Ezra (Abraham Judaeus, Avenare); Zur Geschichte der mathematischen Wissenschaft im XII Jahrhundert; Zeitschr. f└G└r Math. und Physik 25 (1880): Supp: AGM 3, Part II (1880). The material is Art. 20, pp. 123-124. 15 students and 15 goodйforйnothings 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 йй ?? Reinmar von Zweter. Meisterlied: "Ander driu, wie man juden und cristen └E└z zelt". 13C. (In MUS II 128.) Jews and Christians on a ship, counts by 10. Kenk└е└, also known as Kenk└е└ Yoshida or Urabe no Kaneyoshi or Yoshida no Kaneyoshi (1283й1350). Tsurezuregusa. c1331. Translated by Donald Keene as: Essays in Idleness The Tsurezuregusa of Kenk└е└; Columbia Univ. Press, NY, 1967. (Kenk└е└ 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 Mamakoйdate San (or Mamaйko tate no koto й cf Mat└╜└йoka, 1808) or Mamagodate (Scheme to benefit the stepйchildren 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└е└, pp. 115й121 in Keene's version (including a doubleйpage illustration which doesn't depict the problem), whose beginning is characteristic of Kenk└е└'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 йй these are even more deeply moving." The passage of interest is toward the end, on p. 120 of Keene: "When you make a ├├mamagodate──├├1── 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└е└'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 й (e 2).v, which are pages 34й36 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 alйSafadi (Sal└└hadd└3└n as-Safad└3└ = alйS└└phadi =аAlS└└phadi) (d. 1363) in his L└└miyato └└l Agjam (variously printed in the text). This must be his Sharh L└└m└3└yat 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) (8├├o──S.87.Art) where the mnemonic Populeam virgam mater Regina tenebat is given. He then says that the problem is also described in 'Megjdium & Abulphedam' йй p. 43 of his main text identifies Abulpheda as a prince born in 672 AH йй ?? Shih└└badd└3└n Ab└E└└└l-└└Abb└└s Ahmad ibn Yahya ibn Ab└3└ Hajala at-Tilims└└ni alH-anbal└3└ (??sp). Kit└└b └└anm└E└dhaj al-qit└└l fi la└└b ash-shatranj [Book of the examples of warfare in the game of chess]. c1370. Copied by Muhammed ibn └└Ali ibn Muhammed al-Arzag└3└ 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-Safad└3└'s problem of the ship (see Hyde, ii.23)", described by Murray as 15 Christians and 15 Muslims counted by n. Bland has "the wellйknown 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 йй 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.а143. 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 228й230, 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. 102vй103v, pp. 205-207. Coloured plate opp. p.а192 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. ╨ дx ╨╨╨ШМ А th                                                                      д╨╨Ff. 99r й 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 140й143. Does 2 & 30 by 9s йй 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 й 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 й 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 144й145. 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 й 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 LVIIIйLX under one heading which only refers to 'versi memorevili'. ╨ ░x ╨╨╨дШМ А t                                                                      ░╨╨Elias 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 (1494й1576). Meisterleid: 'Historia Die XV Christen und XV T└G└rcken, 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. 303й304. 15 Christians and 15 Jews on a ship counted by 10s. [Mentioned in H&S 52.] Simon Jacob von Coburg. Ein new und Wolgegr└G└ndt 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 йй described in MUS II 133й134. 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. 183й185. 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 йй starting on the place where the counter was removed, so we now would say he is counting four йй 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,а8,а2,а6, 10, 5, 9, 3, 7, 1. The first five values are the values of 4a (mod 10) for aа=а1,а2, 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 й A.7.r; 1884: 8й9 & prob.аXX,а1612:а103-106; prob.аXXIII,аа1624: 174й177; 1884:а118-121. Turks & Christians йй 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. 266й269 (258й261). 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) K└е└y└┴└ (= Mitsuyoshi Yoshida) (1598й1672). Jink└е└-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. 12й14. He gives some of the information on the Japanese names and on Michinori (1106й1159) and Kenk└е└, 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. 66й67. ┴┴┴┴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. 80й84. They quote a slightly later version by Seki K└е└wa (1642й1708) where the stepmother simply reverses the order due to overconfidence. (On p. 121, they identify the source as Sandatsu Kempu, a MS of K└е└wa.) 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 Jinkoйki, cf at 1767. Ahmed elйQalyubi (d. 1659). Naouadir (or Nauadir), c1650?, published at Boulaq (a suburb of Cairo), 1892, hist. 176, p. 82. ??NYS йй described in Basset (1886й1887 below) and MUS II 136. 15 Moslems and 15 infidels on a ship counted by 9s. Muramatsu Kuday└┴└ Mosei. Mantoku Jink└е└-ri. 1665. ??NYS йй described in MUSаIIа139 and Smith & Mikami, pp. 80й84. 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. 531й532. 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 Char└3└gj seu Char└3└tch, pp. 225й226. 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. PetersйSpiel (with a Nachtrag by Bolte). Zeitschrift f└G└r Volkskunde 39 (1929) 190й192 (& 192й194). Schnippel describes the appearance of solutions of the St. PetersйSpiel = Sankt P└└ders Lek = Saint Peter's Lake on 17й18C rune calendars from Sweden, which mystified academics until identified by G. Stephens in 1866. He gives the vowelйmnemonic: Populeam virgam mater regina ferebat. The rune marks are X for └.└└#└└└└%└└)└└└└└└└└└└└ (Xristianoi) and I for └└└└└└└+└└ └└└└Р└└└└└ (Ioudaioi). He gives the German vowelйmnemonic: 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 antiйsemitic folkйtale." 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 1886й1887) and MUS and an article by himself in Euphorion 3 (1896) 351й362, ??NYS йй 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 vowelкmnemonics 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. йй but see below which could explain why Murphy didn't find it here. Ozanam. 1696. Preface to vol. 2 йй first and second of unnumbered pages, which are pp.а269-270. 1708: Author's Preface йй second and third of unnumbered pp. Discusses Josephus, citing Bachet. Ozanam. 1725. Prob. 45, 1725: 246-250. Prob. 17, 1778: 168й171; 1803: 168й171; 1814:а148й150. Prob. 16, 1840: 76й77. 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. 152й154 (1755: 110й111; 1822: 169й171; 1864: 146й148). 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 Jinkoйki (New Edition of the Jonkoйki), more correctly entitled Sanpo Shinan Guruma (A Mathematical Compass). 1767. BL ORB 30/3411. ??NYS йй illustration reproduced in Dean, 1997. Fujita Sadasuke. Sandatsu Kaigi. 1774. ??NYS йй 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. 42й43. 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(nй1) + 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 (= Mat└╜└йoka ??= Matsuoka N└е└ichi). 1808. ??NYS йй 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: Mamaйko tate no koto (Probl└/└me des beauxйfils (i.e. stepкsons)). There is a diagram showing the countingйout 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. 104й106. 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 й 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. 83й85, 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. ╨ дx ╨╨╨ШМ А th                                                                      д╨╨No. 16, pp. 4й5 & 54й56. 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. 9й10 & 61й62. Decimation of a troop of 40 to be counted by 10s йй where to place the four ringleaders so they will be the four to be shot. ╨ ░x ╨╨╨дШМ А t                                                                      ░╨╨Rational 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. 79й80, 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. ╨ дx ╨╨╨ШМ А th                                                                      д╨╨The slighted lady. 1828: 411-412; 1828й2: 417й418; 1829 (US): 210й211; 1855:а565-566; 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; 1828й2: 422; 1855: 571; 1868:а672й673; 1881:а214. 30 criminals counted by 9s to eliminate 15. Populeam virgam mater regina ferebat. ╨ ░x ╨╨╨дШМ А t                                                                      ░╨╨Nuts to Crack XIV (1845), no. 72. 21 counted by 7s to the last man. Magician's Own Book. 1857. ╨ дx ╨╨╨ШМ А th                                                                      д╨╨The fortunate ninth, pp. 221й222. 15 oranges and 15 apples, counted by 9s. English mnemonics based on vowel coding. Another decimation of fruit, p. 224й225. 30 apples and 10 oranges, counted by 12s in order to get the oranges first. ╨ ░x ╨╨╨дШМ А t                                                                      ░╨╨The Sociable. 1858. ╨ дx ╨╨╨ШМ А th                                                                      д╨╨Prob. 31: The puzzle of the Christians and the Turks, pp. 296 & 312й314. 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 & 30й32. Prob. 39: The landlord tricked, pp. 298 & 316. 21 counted by 7s to the last man. =аBook of 500 Puzzles, 1859, prob. 39, pp. 16 & 34. = Wehman; New Book of 200 Puzzles; 1908, p. 51. ╨ ░x ╨╨╨дШМ А t                                                                      ░╨╨The 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. ╨ дx ╨╨╨ШМ А th                                                                      д╨╨The fortunate ninth, pp. 190-191. Identical to Magician's Own Book. Another decimation of fruit, p. 194. Identical to Magician's Own Book. ╨ ░x ╨╨╨дШМ А t                                                                      ░╨╨Vinot. 1860. Art. XXVI: De l'historien Jos└/└phe, pp. 55й56. 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 f└G└r M└└dchen. 1864? Prob. 564й30, 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). CompteйRendu de la premi└/└re session, Paris, 1873. Maisonneuve et Cie., Paris, 1874. T. 1, pp. 289-299, with comments on pp. 299й303. The material of interest is on pp. 294й298. Gives a translation of Mat└╜└йoka, 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. 129й130. 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. 282й285, pp. 50й52 & 100й101; 1895?: 311й314, pp. 54й56 & 102к103; 1917: 311-314, pp. 49-50 & 97-98. ╨ дx ╨╨╨ШМ А th                                                                      д╨╨282 (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 ╨╨╨дШМ А t                                                                      ░╨╨Cassell'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 йй Amusement arithm└)└tique. This is a series of five notes in M└)└lusine 3 (1886-87). ╨ дx ╨╨╨ШМ А th                                                                      д╨╨Gaidoz. Part I. Col. 273-274. Gives classical version with St. Peter, 15 Christians & 15аJews counted by 9s. He then gives two versions from Ceylon. One version is called Yonmaruma йй The massacre of the Moors йй and has 15 Portuguese & 15аMoors 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 йй 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 elйQalyubi, c1650? йй cf above. ╨ ░x ╨╨╨дШМ А t                                                                      ░╨╨Robert Harrison. UK Patent 15,105 йй 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? ╨ дx ╨╨╨ШМ А th                                                                      д╨╨Pp. 105й106: Une dame pas contente. 13 counted by 2s to last person. Pp. 118й119: Stratag└)└me de Jos└)└phe. 41 counted by 3s to last two, claimed to be the method used by Josephus. Pp. 120й121: Les marauders punis. 15 & 15 counted by 9s. Officers and soldiers to be executed. Pp. 121й122: Les naufrages. Same numbers, with Turks and Christians on a boat. Mort,аtu ne failliras pas, En me livrant au tr└)└pas. P. 122: Les └(└lections perfectionn└)└es. 36 counted by 10s йй want the first six chosen. ╨ ░x ╨╨╨дШМ А t                                                                      ░╨╨Hoffmann. 1893. Chap. 4, pp. 156й157 & 210й211 = HoffmannйHordern, pp. 134й135, with photo. ╨ дx ╨╨╨ШМ А th                                                                      д╨╨No. 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, 1874й1895. 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, 1914й1918. ╨ ░x ╨╨╨дШМ А t                                                                      ░╨╨└(└. Lucas. Problem 32. Intermed. des Math. 1 (1894) 9. n├├2── persons, counted by n until n-1 are left. "Probl└/└me dit de Caligula". E. Cesar└A└. 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. 12й18. ╨ дx ╨╨╨ШМ А th                                                                      д╨╨Le stratag└/└me de Jos└/└phe, pp. 12й17. 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. 17й18. 6 and 30 counted by 10s so as to count the 6 first. ╨ ░x ╨╨╨дШМ А t                                                                      ░╨╨H. Schubert. Zw└?└lf Geduldspiele. 1895. P. 125. ??NYS йй 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) 368й372. The prisoners of Omdurman. 8 Europeans followed by 8 Abyssinians in a ring. Start counting with the first European. Determine the countingйout 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 (Julа1900) 303. The "blacks" and "whites" puzzle or The twelve schoolboys. 6аconsecutive "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) 77й90. 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. 83й87 & 90, he reports finding examples in Telugu in the notebooks of the schoolmaster Panakalu Rayudu (1883й1928) 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└]└la└п└iksa, a work which is unknown to me. H. D. Northrop. Popular Pastimes. 1901. No. 8: the landlord tricked, pp. 67й68 & 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 йй 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. =аMPSL2, prob. 42, pp.а30-31 & 134. Like Dudeney's 1905 version with a different arrangement of 5 and 5. Williams. Home Entertainments. 1914. A decimation problem, pp. 122й124. 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. 26й27. 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. 70й71. 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 йй 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. 102й103, pp. 37й38 & 98. ╨ дx ╨╨╨ШМ А th                                                                      д╨╨No. 102. 13 counted by 9s until last man. No. 103. 17 and 15 counted by 12s to eliminate the 15 first. ╨ ░x ╨╨╨дШМ А t                                                                      ░╨╨Ern Shaw. The Pocket Brains Trust йй No. 2. Op. cit. in 5.E. c1944. Prob. 50: Poser with pennies. Pattern of 5 Hs and 5 Ts given йй 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. 111й114. 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 йй 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) 87й93. A more detailed description than in the Summary appears in Rankin's review in: Math. Reviews 10, #509b (= A99й5 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 йй but two abridged versions have appeared which contain the material йй 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 nonйface 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 йй Discarding cards. AMM 57 (1950) 34й35 & 488й489. 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 rйth card is discarded and gets the recurrence f(N)а└└аrа+аf(N-1)а(modаN). 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 BrainйTwisters. 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) 168й173. Gives a method of determining when the iйth man is removed and which is the kйth to be removed. Somewhat similar to Rankin's method. D. Woodhouse The extended Josephus problem. Revista Matematica HispanoйAmericana 33 (1973) 207й218. 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. 121й128. They study the permutation where f(i) = number of iйth 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) 97й104. 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 └└а= 1.216703..., and obtains a formula for L(n). Presumably this could be extended to the general case?? Michael Dean. Josephus and the ├├Mamakoйdate san── (Scheme to benefit the stepйchildren). International Netsuke Society Journal 17:2 (Summer 1997) 41й53. 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) йй see above at 1767, 1795, 1818 йй 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) 30й33. 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 eйth man counted out, form a sequence starting with 1 + k(nйe) and then form each next term by multiplying a term by k/(kй1) and rounding the result up to an integer. (I.e. x├├n+1── = └I└x├├n── * k/(kй1)└J└.) Then the position number of the eйth 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 eа=аn, 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(nйe) + 1 to eliminate the eйth 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, └K└a├├n──└L└. 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─ ─ ╨╨░дШМ А t                                                                    ░╨╨┴┴The 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, cй1825. ╨╨дШМ А t                                                                      ░╨╨ Papyrus Rhind, composed cй1785 (or cй1825), copied c-1650 (or cй1700). A. B. Chace, ed. (1927-29); c=аNCTM, 1978. Pp. 21-22, 50-51. Moscow Mathematical Papyrus. cй1785. W. W. Struve, ed; Mathematischer Papyrus des Staatlichen Museums der Sch└?└nen K└G└nste in Moskau; Quellen und Studien zur Geschichte der Mathematik, Abt. A: Quellen, Band 1; Springer, 1930. Fibonacci. 1202. Pp. 77-83 (S: 119й126): ... 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 (1977й78) 81й86. 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 (cй2134); Rhind Papyrus (cй1785); Moscow Papyrus (cй1785); Kahun Papyri (cй1785, but later than the previous two items); Egyptian Mathematical Leather Roll (cй1647), 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.а329) (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─ ─ ╨╨░дШМ А t                                                                    ░╨╨┴┴Most 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,а1888; Ball, 1890; Clark, 1904; and Pearson, 1907, qv. ┴┴NOTATION. The classic coconuts problem has the following recurrence for the number of coconuts remaining: ┴┴┴┴A├├i+1── = (nй1)/n [A├├i── й 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. ╨ дx ╨╨╨ШМ А th                                                                      д╨╨Ending 0 йй 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 йй the n-th man leaves one more than a multiple of n, so the monkey gets another coconut. See: CarrollйWakeling; Ball; Clark; Pearson; Roray; Collins; Kraitchik; Phillips; Home Book; Leeming; Devi; Allen. One can extend this to Ending E йй the nйth man leaves a number └└ E (mod n). ╨ ░x ╨╨╨дШМ А t                                                                      ░╨╨Other indeterminate versions: Ozanam; Dudeney; Weber (Dirac); Rudin. For the solution with й(nй1) 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 ┴┴┴┴A├├i+1──аа=аа(n-1)A├├i──/n й 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 A├├i── └└ 1 (mod n), and in Morris's form, each A├├i── └└ 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 nйth 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. cй1700?. Transcribed, translated and commented on in: O.аNeugebauer & A. Sachs; Mathematical Cuneiform Texts; American Oriental Society and American Schools of Oriental Research, New Haven, 1945, pp. 100й103, 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." ╨ дx ╨╨╨ШМ А th                                                                      д╨╨No. 7 is reconstructed as: I found a stone, but did not weigh it; after I added oneкseventh and added oneйeleventh, I weighed it: 1 maйna. 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 й x/7 + (x й x/7) / 11 = 1. No. 9 leads to x й x/7 + (x й x/7) / 11 й [x й x/7 + (x й 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 й x/6 + (x й x/6) / 24 = 1. ╨ ░x ╨╨╨дШМ А t                                                                      ░╨╨Old Babylonian tablet YBC 4669. cй1700?. Neugebauer and Sachs continue on p. 103 with a new analysis of this table which Neugebauer had previously treated in Mathematische Keilschriftйtexte 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. ╨ дx ╨╨╨ШМ А th                                                                      д╨╨Chap. 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 ╨╨╨дШМ А t                                                                      ░╨╨Zhang 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 f└G└r d. deutsch└?└sterr. Gymnasien 69 (1919) 112й117. 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 йй 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). ╨ дx ╨╨╨ШМ А th                                                                      д╨╨Prob. 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: 21└E└. 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 йй 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 ╨╨╨дШМ А t                                                                      ░╨╨Papyrus of Akhmim. c7C. Jules Baillet, ed. Le Papyrus Math└)└matique d'Akhm└3└m. 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. 58й59. ╨ дx ╨╨╨ШМ А th                                                                      д╨╨Prob. 13, p. 70. Take 1/13th, then 1/17th of the rest, leaving 150. Answer:аа172а+а1/2а+ 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:а 224а+а1/4 + 1/18. ╨ ░x ╨╨╨дШМ А t                                                                      ░╨╨Bakhshali 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 йй merchant pays customs of 1/3,аа1/4, 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. 52rй52v) or a result of 2x й 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 йй but above we have seen both kinds. Datta, p. 46, says (Kaye III 184,) f.а70v 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 86а1/3 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 Libu└▒└e, in the 7C or 8C, fell into a trance and told her followers to seek a ploughman with two oxen. Such a man, named P└л└emysl (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. 129й140, pp. 67й69 are simple problems of this general type, involving sums of numbers diminished by fractions йй I give just some examples. Chap.аVI, v. 112, 114, 130, 131, 132, pp. 116 & 123-125. Chap. III. ╨ дx ╨╨╨ШМ А th                                                                      д╨╨133. 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/2а+ y/3 + 3z/8 = 1/2 and he arbitrarily picks two of the values, getting: 1/3,а1/4, 2/3. 134. x(1/2)(5/6)(4/5)(7/8)(6/7) = 1/6. ╨ ░x ╨╨╨дШМ А t                                                                      ░╨╨ Chap. VI. ╨ дx ╨╨╨ШМ А th                                                                      д╨╨112. 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 йй (subtract 1 and halve) twice, leaving some even number йй i.e. Ending 0 with 2 men. Cf Pearson, 1907. 132. Man placing flowers in a temple йй (subtract 1 and delete └@└) thrice, leaving some multiple of 3 йй 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 ╨╨╨дШМ А t                                                                      ░╨╨Chaturveda. 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 alйmu└└└]└mal└]└t. c1075. ??NYS. ╨ дx ╨╨╨ШМ А th                                                                      д╨╨Pp. 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 ╨╨╨дШМ А t                                                                      ░╨╨Abraham. 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. 304й376, Paris, 1838. ??NYR йй 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: 372й383, 397й398, 439й445, 460й461) gives many versions! Chap. 12, part 6, pp. 258-267 (S: 372й383): De viagiorum propositionibus, atque eorum similium [On problems of travellers and also similar problems] is devoted to such problems. ╨ дx ╨╨╨ШМ А th                                                                      д╨╨P. 258 (S: 372й373). (Double & spend 12) thrice to leave 0 or 9. Answers:аа10└└,аа11└D└. 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 11└D└ and leaving 9. P. 259 (S: 374й375). (Triple and spend 18) four times to leave 0. Answer: 8а8/9. P. 260 (S: 375). Start with 8 8/9, (triple and spend x) four times to leave 0. Pp. 260-261 (S: 375й376). (Triple and spend 18) four times to leave 12, or the original amount, or original amount + 20. P. 261 (S: 376й377). 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: 377й383). 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 log├├2──а14а=а3.80735. P. 278 (S: 397й398). 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: 439й443). 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 100а(.9)├├12── using a form of decimal fraction, getting 28.2429536481. See 7.L for related problems. Pp. 316-318 (S: 443й445). 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 iа=а2, ..., 10, leaving 1. P. 329 (S: 460). Same as on p. 258, done by false position. P. 329 (S: 460й461). Start with 12, (double and spend x) thrice to leave 0. ╨ ░x ╨╨╨дШМ А t                                                                      ░╨╨Abbot Albert. c1240. Prob. 8, p. 334. (Double and subtract 1) thrice, leaving 0. Chu Shih-Chieh (= Zhu Shijie). Ssu Yuan Y└G└ 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. ╨ дx ╨╨╨ШМ А th                                                                      д╨╨No. 89, pp. 108-109. (Double and spend 35) thrice leaving 0. No. 119, pp. 134-135. (Double and spend 40) thrice leaving 0. ╨ ░x ╨╨╨дШМ А t                                                                      ░╨╨Folkerts. Aufgabensammlungen. 13й15C. ╨ дx ╨╨╨ШМ А th                                                                      д╨╨(Double and give a) n times to leave nothing. 17 sources. Folkerts notes the solution is a й a/2├├n── 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 ╨╨╨дШМ А t                                                                      ░╨╨Gherardi?. Liber habaci. c1310. Pp. 144-145. Three porters йй i-th takes half plus i, leaving none. Gherardi. Libro di Ragioni. 1328. ╨ дx ╨╨╨ШМ А th                                                                      д╨╨Pp. 47-48. Man gathering apples. Four porters йй 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 ╨╨╨дШМ А t                                                                      ░╨╨Lucca 1754. c1330. ╨ дx ╨╨╨ШМ А th                                                                      д╨╨Ff. 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 ╨╨╨дШМ А t                                                                      ░╨╨Paolo 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 йй described in Rara, 435-440 and Van Egmond's Catalog 254й255. Van Egmond 365 lists 9 MSS of this work. MSаBа2433, Biblioteca Universitaria, Bologna, is a c1513 copy of just the problems of this work йй 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 67й68. ┴┴┴┴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. 25rй25v of B 2433. (Halve and subtract 1) thrice to leave 3. Munich 14684. 14C. ╨ дx ╨╨╨ШМ А th                                                                      д╨╨Prob. V, p. 78. (Double and subtract 2) some times to leave 0 йй determines initial values for various numbers of times as 2(1ай 1/2├├n──). 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 ╨╨╨дШМ А t                                                                      ░╨╨Bartoli. 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) 26й75. 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. ╨ дx ╨╨╨ШМ А th                                                                      д╨╨P. 58. Man doubles his money and spends 1, triples and spends 2, quadruples and spends 2, leaving him with 3. F. 113vй114r, 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 ╨╨╨дШМ А t                                                                      ░╨╨Pseudoйdell'Abbaco. c1440. ╨ дx ╨╨╨ШМ А th                                                                      д╨╨Prob. 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 ╨╨╨дШМ А t                                                                      ░╨╨AR. c1450. No. 185 & 187, pp. 87-88, 173-174 & 220. ╨ дx ╨╨╨ШМ А th                                                                      д╨╨185 = Fibonacci, p. 258. 187. Double and spend 6, double and spend 12, double and spend 15, leaving the initial amount. ╨ ░x ╨╨╨дШМ А t                                                                      ░╨╨Muscarello. 1478. ╨ дx ╨╨╨ШМ А th                                                                      д╨╨Ff. 78vй79r, pp. 194й196. Lose 1/2 and 6 й i more, for i = 1, 2, 3, 4, 5, leaving 1. Ff. 84rй85r, pp. 201й204. Merchant starts with 79 and makes profits of 17%, 19%, 21%, 23% at four fairs. ╨ ░x ╨╨╨дШМ А t                                                                      ░╨╨della Francesca. Trattato. c1480. ╨ дx ╨╨╨ШМ А th                                                                      д╨╨F. 23r (73й74). Gain 1/3 + 1/4 and 20 more. Then spend 1/4 + 1/5 and 20 more to leave 24. F. 37v (97й98). 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 ╨╨╨дШМ А t                                                                      ░╨╨Chuquet. 1484. ╨ дx ╨╨╨ШМ А th                                                                      д╨╨Prob. 30. (Double and subtract 12) thrice, leaving 0. English in FHM 206. Prob. 31й33 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 iйth 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 й 9/(x+12). Chuquet uses some interpolation to estimate X = └└(50 16297/16384) й 4 13/128 [FHM misprints this] = 3.03949414, but I get 3.045827298. Chuquet says ordinary interpolation is not valid. ╨ ░x ╨╨╨дШМ А t                                                                      ░╨╨Calandri. Arimethrica. 1491. ╨ дx ╨╨╨ШМ А th                                                                      д╨╨F. 66v. (Double and spend 2) thrice leaving 0. F. 74r. Double and then gain 50% giving 1000. Woodcut of merchant on horse. ╨ ░x ╨╨╨дШМ А t                                                                      ░╨╨Pacioli. Summa. 1494. ╨ дx ╨╨╨ШМ А th                                                                      д╨╨F. 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 2├├x── = 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. 187rй187v, 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── = x├├2──. 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 2├├x── = 14, but again he has to interpolate on the third day. ╨ ░x ╨╨╨дШМ А t                                                                      ░╨╨Calandri, Raccolta. c1495. Prob. 16, pp. 17-18. Merchant gains └@└ of his money plus i on the iйth trip. After three trips he has 15. Pacioli. De Viribus. c1500. Ff. 120r й 120v, 111r й 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 156й158. (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 й 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.а89r, ??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 .9├├10──. (H&S 59й60). Tonstall. De Arte Supputandi. 1522. ╨ дx ╨╨╨ШМ А th                                                                      д╨╨Quest. 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. 173й174. Give half and get back 2i for iа=а1,а2,а3, leaving 12. P. 246. (Double and spend 12) thrice to leave 0. ╨ ░x ╨╨╨дШМ А t                                                                      ░╨╨Riese. Die Coss. 1524. Several examples йй no. 35, 53, 55, 56, 58, 59, 60, 61, 62. I describe a few. ╨ дx ╨╨╨ШМ А th                                                                      д╨╨No. 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 ╨╨╨дШМ А t                                                                      ░╨╨Tartaglia. General Trattato. 1556. Book 12, art. 34, p. 199v. Book 16, art. 47 & 113й116, pp.а246r & 253v-254r. Book 17, art. 9 & 20, p. 268v & 271r. Final remainder specified in each case. ╨ дx ╨╨╨ШМ А th                                                                      д╨╨12й34. (Take half plus i more) for i = 1, 2, 3, 4, leaving 1. Cf 16й115. 16й47. Take 1/2 and 1 more, 1/3 and 2 more, 1/4 and 4 more, leaving 26. 16й113. (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). 16й114. (Halve and subtract 1) thrice, leaving 1, 2, .... 16й115. Halve and subtract 1, then 2, 3, 4, leaving 1. Cf 12й34. 16й116. 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. 17й9. Double & spend 4, double & spend 8, leaving 24. 17й20. Double and spend 18, double and spend 24, double and spend 36, leaving 280. ╨ ░x ╨╨╨дШМ А t                                                                      ░╨╨Buteo. Logistica. 1559. ╨ дx ╨╨╨ШМ А th                                                                      д╨╨Prob. 6, pp. 334й335. Lose 1/2 and 3 more, lose 1/3 and 4 more, lose 1/4 and gain 1, to leave 100. Prob. 13, pp. 342й343. 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. 347й348. Gain 1/4 and spend 7, gain 1/3 and spend 10, lose 3/7 and spend 8, leaving 0. Prob. 21, pp. 348й350. (Double and spend 10) X times to leave 0. He makes an error at X = 8 and deduces X = 7. ╨ ░x ╨╨╨дШМ А t                                                                      ░╨╨Baker. Well Spring of Sciences. 1562? ╨ дx ╨╨╨ШМ А th                                                                      д╨╨Prob. 7, 1580?: ff. 192rй193r; 1646: pp. 302-304; 1670: pp. 344й345. Lose half and gain 12, lose half and gain 7, lose half and gain 4, leaving 20. Prob. 8, 1580?: ff. 193rй193v; 1646: p. 304; 1670: p. 345. (Double and spend 10) thrice leaving 12. ╨ ░x ╨╨╨дШМ А t                                                                      ░╨╨Gori. 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 2├├k── if one does the process k times. Montucla replaces this by some determinate problems йй see below. Simpson. Algebra. 1745. Section XI (misprinted IX in 1790), prob. XV, pp. 86й87 (1790: prob. XXVII, pp. 88й89. Shepherd loses (half and 1/2 more) thrice to leave 5 (or 1) sheep. Les Amusemens. 1749. ╨ дx ╨╨╨ШМ А th                                                                      д╨╨Prob. 108, p. 249. (Double and give away 6) thrice to leave 0. Prob. 111, pp. 252й253. 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 ╨╨╨дШМ А t                                                                      ░╨╨Walkingame. Tutor's Assistant. 1751. ╨ дx ╨╨╨