16-120. 40 for 480 at 36, 12, 1. (Curlews or siskins, stock-doves, starlings.) (2, 1) solutions, he gives (1, 1). 16-121. 40 for 40 at 3, 2, 1/5. (Blackbirds, larks, sparrows.) (1, 1) solutions, which he gives. 16-122. 31 for 31 at 3, 2, ⅓. (Capons, ducks, thrushes.) (1, 1) solutions, he gives (1, 1). 16-123. 100 for 100 at 3, 1, 1/20. (Piglets, goats, weasels??) (2, 1) solutions, he gives (1, 1). = Lucca 1754, p. 10r. 16-124. 60 for 60 at 3, 1, 1/20. (Same? animals.) (2, 1) solutions, he gives (1, 1). 16-125. 100 for 100 at 3, 2, 1/20. (Same? animals.) (1, 1) solutions, which he gives. (Probs. 126 129 involve men, women, children eating.) 16-126. 12 for 12 at 2, ½, ¼. (2, 1) solutions, he gives (1, 1). = Alcuin 47. 16-127. 15 for 15 at 4/3, 2/3, 1/3. (I read the last 3 as a 2, but the answer implies it is a 3.) (3, 2) solutions, he gives (1, 1). 16-128. 18 for 18 at 2, 1, ½. (7, 5) solutions, he gives (1, 1). 16-129. 20 for 20 at 4, ½, ¼. (2, 1) solutions, he gives (1, 1). = Pacioli 18. 17-25. 16 for 640 at 32, 50. (Two types of cloth.) This is determinate and he gives the solution. 17-26. 6 for 332 at 42, 66. (Two types of cloth.) Determinate -- he gives the solution. 17 -- 44-45. See also Bachet, below. 17-44. 100 for 100 at 3, 1, ½, ⅓. (Asses, pigs, sheep, goats.) Quoted from Luca Pacioli, p. 105. (276, 226) solutions -- he gives one but notes that the method of double false position does not totally resolve the problem, but that it can be solved by a combination of trying and of mathematics, but this is too long to describe here and he reserves it for another time. = Pacioli 17 17-45. 200 for 200 at 12, 3, 1, ½, ⅓. (Mules, asses, pigs, goats, sheep.) "Proposed to me in 1533 by a Genovese." (8331, 6627) solutions -- he gives one and indicates that more are possible, saying again that he will deal with this at another time.
Buteo. Logistica. 1559.
Prob. 66, p. 274. 50 for 160 at 7, 2, 4, 1. (Partridges, thrushes, quails, fig-peckers.) Cites Pacioli for similar problems. Gives (2, 2) of the (163, 144) solutions. Prob. 67, pp. 274-275. 3x + 9y + 2z = 50; 7x + 3y + 6z = 70. Cites Pacioli for a similar problem. This has no non-negative integral solutions. Gives three positive solutions, choosing different values to be integral.
Baker. Well Spring of Sciences. 1562? Prob. 11, 1580?: ff. 194r-195r; 1646: pp. 305 306; 1670: pp. 346-347. 20 for 240 at 20, 15, 8. (Payments to men, women, children.) (1, 1) solutions, which he gives.
Bachet. Problemes. 1612. Addl. prob. X, 1612: 164-172; 1624: 237-247; 1884: 172 179.
1612 cites Tartaglia, Pacioli, de la Roche, etc. 41 for 40 at 4, 3, ⅓. (1, 1) solution which he gives. 20 for 20 at 4, ½, ¼. (2, 1) solutions -- he gives (1, 1). = Pacioli 18. He then describes the last two examples of Tartaglia. For art. 44, he says there are 226 solutions and gives some. For art. 45, my copy of Bachet has a defective type which made me think that the price of 3 was a 5, but seeing Tartaglia has corrected this error. Bachet says this has 6639 solutions and he gives the numbers for each given number of mules. I find that for 7 mules, he has 571 instead of 570 and for 4 mules, he has 914 instead of 903, which accounts for his extraneous numbers, but I cannot see why he might have miscounted. Labosne adds a general argument for art. 44.
Book of Merry Riddles. 1629? 12 for 12 at 4, 2, ½, ¼. (Capons, hens, woodcocks, larks). (5, 1) solutions, with the positive one given.
John Wallis. A Treatise of Algebra, both Historical and Practical. John Playford for Richard Davis, Oxford, 1685. (Not = De Algebra Tractatus.) Chap. LVIII, pp. 216-218. ??NX
20 for 20 at 4, ½, ¼. (Geese, quails, larks.) = Pacioli 18. He gives the positive solution of the (2, 1) solutions. Cites Bachet and gives some general discussion. 100 for 100 at 3, 1, ½, 1/7. Says the solutions are in Bachet, but they are not. (121, 81) solutions.
W. Leybourn. Pleasure with Profit. 1694. Chap. XIII: Of Ceres and Virginum, pp. 51-55.
Quest. 1, p. 51. 8 for 20 at 4, 2. (Geese, hens). This is determinate. Quest. 2, pp. 51-53. 21 for 26 at 2, 1, ½. (Men, women, children.) Some general discussion. From x + y + z = N and ax + by + cz = P, assuming a > b > c, he gets (a c)x + (b-c)y = P - cN and deduces that x (P-cN)/(a-c). He then assumes x y and hence that x cannot be much less than (P-cN)/(a-c+b-c) and he later drops the 'much' from this. There are (6, 5) solutions; he finds all of them, but rejects the case 5, 16, 0 as the problem says there are 'some children'. Quest 3, p. 53. 30 for 900 at 60, 40, 20. (Ministers, lame soldiers, poor tradesmen.) From the above argument, he claims that x 3, but then gives the solution 2, 11, 17 as though this showed that x = 2 was impossible. He finds (6, 6) of the (8, 7) solutions. Quest 4, p. 54. 10 for 1000 at 50, 70, 130, 150. (English, Dutch, French, Spanish) creditors. Finds (2, 2) of the (10, 4) solutions and implies that all solutions are symmetric. Quest 5, p. 55. 12 for 12 at 2, 1, ½, ¼. (Different prices of loaves of bread.) He finds (2, 2) of the (11, 4) solutions.
Anonymous proposer and solver. Ladies' Diary, 1709-10 = T. Leybourn, I: 5, quest. 8. Mix wines worth 32, 20, 16 to make 56 worth 22. I.e. 56 for 1232 at 32, 20, 16. (12, 11) solutions of which the positive ones are given.
Adrastea, proposer; anonymous solver. Ladies' Diary, 1721-22 = T. Leybourn, I: 112-113, quest. 89. 24 passengers of four ranks pay £24, with their fares in the proportion 16 : 8 : 2 : 1. The solutions depend on whether one takes fares as whole numbers of pounds, shillings or pence. The cheapest fare, R, is readily seen to satisfy 15d R 1£. The solution says a Mr. Evans collected 100 true answers. I found (1, 0) solutions in pounds and (89, 43) solutions in shillings, then wrote a special program to solve the 24 cases which occur in pence, finding (201, 101) solutions. Presumably Mr. Evans missed one of the positive solutions. Using farthings gives just one more case with (8, 3) solutions.
Ozanam. 1725. [This is one of the few topics where the 1725, 1778 and 1803 editions vary widely -- see each of them.]
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