Example 2 (VII 11). 20 for 20 at ?, 4, ½. From the working, the lost coefficient satisfies 0 ? 15/32. One solution is given, but only partly readable: ?, 3, ?. Not mentioned by Kaye. Example 3 (VII 11-12). 20 for 20 at 3, 3/2, ½. (Earnings of men, women, children.) (3, 1) solutions, (1, 1) given: 2, 5, 13. (Also in Kaye I 42; III 191, f. 58v. A. K. Bag; Mathematics in Ancient and Medieval India; 1979, p. 92 gives just the problem. Datta, p. 50, says the answer is mutilated, but Hayashi does not comment on this -- Kaye III 191 displays the answer given as 2, 5, 1..., so the mutilation is pretty minimal.)
Li Shun fêng. Commentary on Chang chiu chien. 7C. ??NYS. Libbrecht (p. 280) describes his comments as "unmitigated nonsense".
L. Vanhée. Les cent volailles ou l'analyse indéterminée en Chine. T'oung Pao 14 (1913) 203 210 & 435 450. On pp. 204 210, he gives 24 problems in Chinese and French, but doesn't identify the sources!!
Alcuin. 9C.
Prob. 5: Propositio de emptore in C denarius. 100 for 100 at 10, 5, ½. (Boars, sows, piglets.) (1, 1) solution, which he gives. Prob. 32: Propositio de quodam patrefamilias distribuente annonam. 20 for 20 at 3, 2, ½. (Dividing grain among men, women, children.) This has (2, 1) solutions -- he gives (1, 1). Prob. 33: Alia propositio. 30 for 30 at 3, 2, ½. (Like Prop. 32.) (3, 1) solutions, (1, 1) given. Prob. 33a (in the Bede text): Item alia propositio. 90 for 90 at 3, 2, ½. (Like Prob. 32.) (7, 5) solutions, (1, 1) given. Prob. 34: Item alia propositio. 100 for 100 at 3, 2, ½. (Like Prop. 32.) (7, 6) solutions, (1, 1) given. Prob. 38: Propositio de quodam emptore in animalibus centum. 100 for 100 at 3, 1, 1/24. (Horses, cows, sheep.) (2, 1) solutions, (1, 1) given. Prob. 39: Propositio de quodam emptore in oriente. 100 for 100 at 5, 1, 1/20. (Camels, asses, sheep.) (2, 1) solutions, (1, 1) given. Prob. 47: Propositio de episcopo qui jussit XII panes in clero dividi. 12 for 12 at 2, ½, ¼. (Dividing loaves among priests, deacons, readers.) (2, 1) solutions, (1, 1) given. Prob. 53: Propositio de homine patrefamilias monasteri XII monachorum. This problem seems to be a major corruption of the following. 12 for 204 at 32, 8, 4. (Dividing eggs among priests, deacons, readers.) This problem has one solution: 5, 4, 3 and it seems like the problem was last and some scribe has used the result to reformulate the problem as giving 204/12 = 17 to each.
Mahavira. 850. Chap. III, v. 133, pp. 67-68 is related to this general type. Chap VI, v. 143 153, pp. 130 135.
Chap. III.
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 is: x/2 + y/3 + 3z/8 = 1/2 and he arbitrarily picks two of the values, getting: 1/3, 1/4, 2/3.
Chap. VI.
143: complex triple version with prices also to be found. 146: general method. 147: 72 for 56 at ⅔, ¾, 4/5, 5/6. (Peacocks, pigeons, swans, sârasa birds.) (217, 169) solutions, (1, 1) is given. 150: 68 for 60 at 3/5, 4/11, 8. (Ginger, long pepper, pepper.) (1, 1) solutions, (1, 1) given. 151: gives a complex method. 152: 100 for 100 at 3/5, 5/7, 7/9, 9/3. (Pigeons, sârasa birds, swans, peacocks.) (26, 16) solutions, (1, 1) given. (See Sridhara & Bhaskara II below.)
Pseudo-Alcuin. 9C. ??NYS -- cited by Hermelink, op. cit. in 3.A.
Sridhara. c900. V. 63 64, ex. 78 80, pp. 50 52 & 95.
Ex. 80. 100 for 80 at (2, 3/5, ½). (Pomegranates, mangoes, wood apples.) Commentator gives (5, 5) solutions of the (6, 5).
Abu Kamil [Abū Kāmil Shujā‘ ibn Aslam ibn Muhammad (the h should have an underdot) ibn Shujā‘, al-Hāsib al-Mişrī]. Kitāb al ţara’if [NOTE: ş, ţ denote s, t with an underdot.] fi’l hisāb (the h should have an underdot) (Book of Rare Things in the Art of Calculation). c900. Trans. by H. Suter as: Das Buch der Seltenheiten der Rechenkunst von Abū Kāmil el Mişrī; Bibl. Math. (3) 11 (1910 1911) 100 120. (I have a reference to an Italian translation by G. Sacerdote; IN: Festschrift zum 80 Geburtstag M. Steinschneiders; Leipzig, 1896, pp. 169 194, ??NYS.) (Part is in English in Ore; Number Theory and Its History; 139 140.) Six problems of 100 for 100 at the following.
1. 5, 1, 1/20. (Ducks, hens, sparrows.) (2, 1) solutions, (1, 1) given. 2. 2, ⅓, ½. (Ducks, doves, hens.) (7, 6) solutions, he gives (6, 6). 3. 4, 1/10, ½, 1. (Ducks, sparrows, doves, hens.) (122, 98) solutions, he gives (96, 96) and an early commentator pointed out the missing two positive solutions. 4. 2, ½, ⅓, 1. (Ducks, doves, larks, hens.) (364, 304) solutions, he says (304, 304). 5. 3, ⅓, 1/20. (Ducks, hens, sparrows.) (1, 0) solutions, he says there is no solution. 6. 2, ½, ⅓, ¼, 1. (Ducks, doves, ring doves, larks, sparrows.) (3727, 2678) solutions -- he says 2676 once and 2696 twice. Suter notes that the Arabic words for 70 and 90 are easily confused. Suter's comments say there are 2676 solutions. Tom O'Beirne [Puzzles and Paradoxes; OUP, 1965; Chap. 12] discusses this and finds 2678 solutions -- he is apparently the first to find this number, but this probably appeared in his New Scientist column in 1960-1961, ??NYR. The Appendix is a fragment of a commentary which the translator fills in to be Prob. A1: 400 for 400 at 1, 3, 2, 1/7. (Doves, partridges, hens, sparrows.) This has (1886, 1806) solutions. Suter estimates c1700. The commentator and Suter only check the case when the number of doves is divisible by 5. This has (398, 342) solutions. The commentator mentions 334 solutions and Suter says 341.
al Karkhi. c1010. Sect II, no. 10, p. 82. Mix goods worth 5, 7, 9 to make one worth 8.
Tabari. Miftāh al-mu‘āmalāt. c1075. P. 110f., part IV, no. 20. ??NYS - cited by Tropfke 614, who says it has (cows, sheep, hens).
Bhaskara II. Bijaganita. 1150. Chap VI, v. 158 159. In Colebrooke, pp. 233 235. Same as Mahavira's 152. (Doves, cranes, geese, peacocks.) (This also has 5x + 8y + 7z + 92 = 7x + 9y + 6z + 62. ??)
Fibonacci. 1202. Chap. 11: de consolamine monetarum [on the alloying of monies], pp. 143 166 (S: 227-257) deals with this problem in the more general form of combining metals. (Other versions involve mixing spices, wines, etc.) These lead to a1x1 + ... + anxn = b(x1 + ... + xn), often with x1 + ... + xn specified. xi is the weight and ai is the purity or gold content, etc., of the i-th metal. We are mixing the metals to produce a total of weight x1 + ... + xn with purity b. Hence there is no need to consider only integral values but he usually gives one (or a few) integral solutions. I describe a few examples. A denotes the vector of ai's.
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