7. arithmetic & number theoretic recreations a. Fibonacci numbers
III-(120, 180, 240, 300, 360). Answer: 180, 120, 60, 0, -60. Sesiano notes these numbers occur in the solution of the problem on p. 641 -- see 7.R.2
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III-(120, 180, 240, 300, 360). Answer: 180, 120, 60, 0, -60. Sesiano notes these numbers occur in the solution of the problem on p. 641 -- see 7.R.2.FHM 83 gives the next four just by formulae with solutions.II-(2, 3, 4) with purse worth 40. Answer: (280, 680, 920)/73.I-(3, 4, 5, 6) with purse worth 50. Answer: (200, 3350, 5450, 6950)/941.I-(2, 3, 4, 5) with purse worth 26. Answer: ( 78, 312, 546, 702)/123. FHM puts down +78 and hence misses this interesting example which Sesiano discusses.Version giving equations like w + z = 2 (x + y) = 26⅔, but this is a bit far away from the problems of this section.FHM 83-84: Indeterminate version of the Bloom of Thymarides: w + x = 17, w + y = 18, w + z = 19. He solves by setting w = 12, and says any other value less than 17 can be used "wherefor such calculations may have innumerable responses."Appendice. Prob. 81. III-(19, 23, 30). Answer: 17, 13, 6.Prob. 82, English in FHM 212-213. (3, 5). Answers: 4, 6; 14 and (60, 90; 210)/7. Says "such questions do not have a necessary answer."Borghi. Arithmetica. 1491? -- this material is additional to the 1484 ed. and Rara indicates that the first ed. to have 100ff is the 4th of 1491. The folio numbers are from the 1509 ed. Ff. 99r-99v. Men find a purse, I-(3, 4). = Fibonacci p. 212.Ff. 99v-100r. Men find a purse, I-(2, 3, 4). = Fibonacci p. 213. = Iamblichus' 1st.F. 100r. I-(3, 4) with purse worth 16.F. 100r. I-(2, 3, 4) with purse worth 30.Calandri. Arimethrica. 1491. F. 65v. III-(12, 14, 10).F. 65v. I-(6, 40). Answer: 7, 41; 239.Pacioli. Summa. 1494. Some of his problems mix this with 7.R.2. F. 105v, prob. 19. Two men find two purses of values p+10, p, giving equations: x + p+10 + 10 = 4 (y - 10), y + p + 20 = 5 (x - 20). He assumes the purses are worth 100 in total, so p = 45, p+10 = 55. Answer: (765, 690)/19. = Tonstall, pp. 245-246.F. 190v, prob. 22. I-(3, 4). Answer: 4, 5; 11.Ff. 190v-191r, prob. 23. x + p = 2y + 2, y + p = 4x + 2. Answer: 12, 20; 30.Ff. 192r-192v, prob. 28. I-(2, 3, 4). He assumes p = 10 and gets answer: (70, 170, 230; 730)/73.F. 192v, prob. 30. I-(3/2, 7/3, 15/4). Assumes p = 1 and gets answer: (63, 177, 279; 621)/621.F. 192v, prob. 31. I-(2, 3, 4). Doesn't observe that this = prob. 28. Answer: (7, 17, 23; 73)/73.F. 193v, prob. 39 & 40. III-(35, 32, 27). III-(122, 114, 106, 96). Says one can deal similarly with more people. Prob. 40 discusses the point further.Ff. 193v-194r, prob. 41. 3 men find a purse and want to buy a horse, giving: x + y + p/3 = h, y + z + p/5 = h, z + x + p/4 = h. If T = x + y + z, one gets T + 47p/60 = 3h and the solution space is actually two dimensional. He assumes p = 60, h = 47 and this problem reduces to prob. 39. Cf della Francesca 40v in 7.R.2.Ghaligai. Practica D'Arithmetica. 1521. He gives several versions, of increasing complexity -- the later ones involve various numbers in geometric progressions or using roots and I omit these. Prob. 13, f. 103r. I-(6, 10). Answer: (42, 66; 354)/7. He arbitrarily sets the first man's money at 6.Prob. 14, ff. 103r-103v. I-(4, 6), but with total money = 100. Answer: (100, 140, 460)/7.Prob. 16, ff. 103v-104r. Two men find purses worth p and p+13, with coefficients 2 and 3, i.e. x + p = 2y, y + p+13 = 3x. Answer: (339, 400; 461)/12.Tonstall. De Arte Supputandi. 1522. Quest. 42, pp. 172-173. III-(150, 240, 326). Considers four person case and sketches general solution.Pp. 245-246. Same as Pacioli's prob. 19, but clearly says the two purses are worth 100.Riese. Die Coss. 1524. No. 45, p. 46. 2 men find a purse, (2, 5).No. 65, p. 49. 2 men find a purse, leading to: x + p + 1 = b 1, y + p + 4 = 3 (x 4). He assumes p = 2, but the general solution is b = 2a 7, p = a 9.Apianus. Kauffmanss Rechnung. 1527. F. M.iii.v. I-(½, 3) with p = 30. Answer: 60, 160.F. M.iv.r. III-(44, 36, 30). Answer: 11, 19, 25.Cardan. Practica Arithmetice. 1539. Chap. 66. Section 97, ff. HH.v.r - HH.vi.r (p. 169). III-(34, 73, 72, 88).Section 98, ff. HH.vi.r - HH.vi.v (p. 169). Men find a purse and buy a horse, giving: x + y + p/2 = y + z + p/5 = x + z + p/3 = h. Answer: 6, 10, 15; 30, 31.Recorde. Second Part. 1552. Pp. 320-322: A question of debt, the third example. IV (47, 88, 71). Answer: 15, 32, 56. Tartaglia. General Trattato. 1556. Book 16, art. 28 35 & 40, pp. 243r 245r. Art. 28. I (2, 3, 4). Answer: 7, 17, 23; 73. (= Iamblichus' 1st.)Art. 29. I (3, 4, 5). Answer: 7, 13, 17; 83.Art. 30. I (4, 6, 8). Answer: 17, 53, 73; 487.Art. 31. Same as 30 with purse given as 100.Art. 32. Same as 30 with total given as 1200.Art. 33. II (2, 3, 4). Answer: 9, 16, 13; 23. (= Fibonacci, p. 216.)Art. 34. II (4, 5, 6). Answer: 25, 36, 31; 119.Art. 35. II (2, 3, 4, 5). Answer: 33, 76, 65, 46; 119. (= Fibonacci, pp. 218 220 & Lucca 1754.)Art. 40. III (24, 28, 32, 36).Buteo. Logistica. 1559. Prob. 9, p. 209-210. III-(4900, 3760, 4660). Remarks on the case with four people. Schott. 1674. Ænigma X, p. 556. Two cups and a cover (equivalent to a purse) worth 90. (2, 3).Ænigma II, p. 563. I-(2, 3, 4) with purses worth 136, 184, 176. Answer: 24, 32, 48.Wells. 1698. No. 117, p. 208. Two horses and equipage: (1, 2) with equipage (equivalent to a purse) worth 5. Simpson. Algebra. 1745. Section XI (misprinted IX in 1790), prob. V, pp. 79-80 (1790: prob. VII, p. 80). III-(17, 16, 15). 1745 gives two methods of solution; 1790 gives one. Mair. 1765? Pp. 458-459, ex. 8. Two horses: (2, 3) with saddle worth 50. Vyse. Tutor's Guide. 1771? Prob. 27, 1793: pp. 56-57; 1799: pp. 61-62 & Key p. 67. III (33000, 30000, 32000, 28000, 25000). Dodson. Math. Repository. 1775.
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