Prob. 10, p. 86. I (½, ⅔) with h = 50. Answer: 37½, 25. Prob. 12, p. 87. x + 3z = 40, z + 2y = 40, y + x = 40. Equivalent to II (3, 2, 1) with h = 40. Answer: x, z, y = (160, 40, 120)/7. Prob. 13, pp. 87 88. Equivalent to II (6, 5, 4, 3, 2). Answer given is 265, 76, 129, 148, 191; 721 and this is the least integral solution.
Diophantos. Arithmetica. c250. Book I.
No. 22, p. 138: "To find three numbers such that, if each give to the next following a given fraction of itself, in order, the results after each has given and taken may be equal." However, he says that the equal amounts are the results after both giving and taking, i.e. (1 b) y + ax = (1 c) z + by = (1 a) x + cz. He does a, b, c = 1/3, 1/4, 1/5. Answer: 6, 4, 5. No. 23, pp. 138 139 is the same with four people and numbers 1/3, 1/4, 1/5, 1/6. Answer: 150, 92, 120, 114. No. 24, p. 139: "To find three numbers such that, if each receives a given fraction of the sum of the other two, the results are all equal." Does I (1/3, 1/4, 1/5). Answer: 13, 17, 19; 25. No. 25, pp. 139 140. Same with four numbers. Does I (1/3, 1/4, 1/5, 1/6). Answer: 47, 77, 92, 101; 137.
Sun Zi. Sun Zi Suan Ching. Op. cit. in 7.P.2. 4C. ??NYS. Chap. III, no. 28. I-(½, ⅔) with h = 48 -- like Chiu Chang Suan Ching, prob. 10. (English in Lam & Shen, HM 16 (1989) 117.)
See 7.P.4 -- Bakhshali MS for a problem which is related, leading to x1/2 + x2 + x3 + x4 + x5 = h, etc., where h is the price of a jewel. Solution: 120, 90, 80, 75, 72; 377. Also an example with three values and diagonal coefficients -7/12, -3/4, -5/6 and solution: 924, 836, 798; 1095.
Sesiano cites Abū Kāmil's Algebra and al-Karajī's Kāfī, ??NYS. Hermelink, op. cit. in 3.A, cites Kāfī & Beha-Eddin, ??NYS
al Karkhi. c1010.
Sect I, no. 42 43, p. 80.
42: x + ⅓y = 20 = y + ¼x. I.e. I-(⅓, ¼) with h = 20. Answer: (160, 180)/11. 43: x + ⅓y + 5 = 20 = y + ¼x + 6. Equivalent to I-(⅓, ¼) with hi = 15, 14. Answer: (124, 123)/11.
Sect. III, no. 26 27, 32 35, pp. 95 100.
26: I (1/3, 1/4, 1/5) with horse worth 20. Answer: (52, 68, 76)/5. See Diophantos I 24. 27: I (1/3, 1/4, 1/5, 1/6). Answer: 47, 77, 92, 101; 137. (= Diophantos I 25.) 32: x = y + ⅓z, y = z + ⅓x, z = x + ⅓y. 33: II (1/3, 1/5, 1/4). Answer: (44, 51, 50; 61)/11. 34: = Diophantos I 22. 35: = Diophantos I 23, but with answer divided by 23, making y = 4.
Tabari. Miftāh al-mu‘āmalāt. c1075. P. 133, no. 52. ??NYS -- Hermelink, op. cit. in 3.A, mentions this without details. Tropfke 609 cites this problem and also p. 150f., no. 13, as buying a horse.
Fibonacci. 1202. Pp. 228 258 (S: 337-372): chap. 12, part 5: De emptione equorum inter consocios secundum datam proportionem [On the purchase of horses among partners according to some given proportion]. Many examples, getting up to seven men, up to five horses, an inconsistent example and negative solutions. I have omitted some of the variations and some of the more complex problems. Some of the problems cited are followed by some discussion.
See: K. Vogel; Zur Geschichte der linearen Gleichungen mit mehreren Unbekannten; Deutsche Mathematik 5 (1940) 217 240; for a thorough study of the problems on pp. 228 258. Some further versions occur on pp. 327 349 (S: 458-484). Vogel cites several sources, ??NYS: Codex lat. Monacensis 14908 (1456) and, for a simpler type of problem, Mich. Pap. 620.
P. 228 (S: 337). I (⅓, ¼). Answer: 8, 9; 11. Cf al-Karkhi I-42. P. 229 (S: 338). II (1/3, 1/4, 1/5). Answer: 45, 48, 52; 61. P. 231 (S: 341). II (2/3, 4/7, 5/9). Answer: 135, 141, 154; 229. Pp. 231 232 (S: 341-342). II (1/3, 1/4, 1/5, 1/6). Answer: 264, 285, 296, 315; 359. Pp. 234 (S: 344-345). II (1/3, 1/4, 1/5, 1/6, 1/7). Answer: 1855, 1998, 2092, 2145, 2156; 2521. The margin has 1815 for 1855 and 2256 for 2156. Pp. 234 235 (S: 345-346). II (2/3, 4/7, 5/11, 6/13, 8/19). Answer: 35435, 35313, 41712, 38643, 44057; 58977. Pp. 235 (S: 346). II (1/4 + 1/3, 1/5 + 1/4, 1/6 + 1/5, 1/7 + 1/6). Answer: 176274, 200772, 205820, 238830; 293391. Pp. 235 236 (S: 346-347). Two men, two horses of values h1, h2 = h, h+2. i th says "If I had ai of what the i+1-st has, then I could buy the i-th horse", with A = (⅓, ¼). I.e. II-(⅓, ¼) with hi = h, h+2. Gives the first two solutions: 8, 12; 12 and 16, 21; 23. Varies to h2 = h+3 and gives the solution 20, 27; 29, which is the third positive solution of the problem. Pp. 236 240 (S: 347-352). Versions with n men and n horses, n = 3, 4, 5. Pp. 240 242 (S: 352-354). Four men, one horse, giving w + ⅓ (x + y) = ... = h, with constants 1/3, 1/4, 1/5, 1/6. Answer: 187, 209, 247, 273; 339. Pp. 242 243 (S: 354-355). Three men, one horse, giving x + y + z/3 = ... = h, with constants 1/3, 1/4, 1/5. Answer: 15, 16, 18; 37. (He interchanges roles of x and y.) P. 243 (S: 355-356). Same with four men and constants 1/3, 1/4, 1/5, 1/6. Answer: 15, 18, 15, 20; 38. The margin has 28 for 38. Pp. 243 244 (S: 356-357). Five men, one horse, giving v + w + x + ¼ y = ... = h. Answer: 1218, 1295, 1200, 1260, 1365; 4028. The margin has 1269 for 1260. P. 245 (S: 357-358). I (1/3, 1/4, 1/5). Answer: 13, 17, 19; 25. (= Diophantos I 24.) Pp. 245 248 (S: 358-360). I (1/2, 1/3, 1/4, 1/5). Answer: 1, 19, 25, 28; 37. Pp. 248 249 (S: 361-362). I (2/5, 3/8, 4/11, 6/19). Answer: 1774, 2047, 2164, 2614; 4504. Margin has 2164 for 2614. States a variation: I (1/3 + 1/4, 1/4 + 1/5, 1/5 + 1/6, 1/6 + 1/7) with answer: 1376, 54272, 76022, 87902; 128657. Pp. 249 250 (S: 362-364): Questio nobis proposita a peritissimo magistro musco constantinopolitano in constantinopoli [A problem proposed to us by a most learned master of a Constantinople mosque]. Buying a ship.
Share with your friends: |
