20 for 20 at 3, 2, ½. (Birds or horses.) 2 sources. (2, 1) solutions, (1, 1) given. = Alcuin 32. 30 for 30 at 2, 1/2, 1/10. (Birds.) 5 sources. (2, 1) solutions, (1, 1) given. Cf AR 45. 30 for 30 at 3/2, 1, 1/2. 1 source. (16, 14) solutions, (1, 1) given. In no case is the solution method given. Folkerts conjectures it was done by double false position. He cites other material, all cited in this section.
Bartoli. Memoriale. c1420. Ff. 89v - 95r (= Sesiano, pp. 134-135). Eleven problems of alligation, many identical to Lucca 1754.
Jamshid al-Kāshī = Ğamšīd ibn Mas‘ūd ibn Mahmūd (the h should have an underdot), Ġijāt ed-dīn al-Kāšī = Ghiyāth al-Din al-Kāshī. Miftāh al-hisāb (the h should have an underdot) (The Calculator's Key). 1426. Ed. by A. Demerdash & M. H. Hifna (the H should have an underdot) ; Cairo, nd. Facsimile and Russian translation by: B. A. Rozenfeld', V. S. Segal' & A. P. Juškevič as: Ključ Arifmetiki -- Traktat ob Okružnosti; Moscow, 1956. ??NYS -- Hermelink, op. cit. in 3.A, says he gives a version with three fowls.
Provençale Arithmétique. c1430. Op. cit. in 7.E.
F. 116r, p. 62. 12 for 12 at 2, 1, ½. (Men, women, children.) (5, 3) solutions, he gives (3, 3). F. 117r, p. 63. 30 for 30 at 4, 2, ½. (Costs of cloth.) Same as Abbot Albert, p. 334. (2, 1) solutions, he gives the positive one. See Chuquet, 1484, prob. 83.
Pseudo-dell'Abbaco. c1440. Prob. 190, pp. 150 151. 48 for 48 at 4, 2, ¼. (1, 1) solution, given.
AR. c1450. Prob. 45, 81, 102, 119 126, 309. Pp. 40, 52, 60 61, 66 67, 137 138, 175 176, 221 222.
45. 30 for 30 at 2, ½, 1/10. (Pears, apples, nuts.) (2, 1) solutions, (1, 1) given. Vogel, p. 221, says this appears in Clm 8951, which is unpublished. = Folkerts, Aufgabensammlungen, sixth example, where Clm 8951 is one of 5 sources cited. 81. Mix three kinds of wax worth 43, 29, 22 to make wax worth 32. 102. Mix wines worth 5, 6, 8 to make wine worth 7. 119. 100 for 100 at 10, 5, ½. (Oxen, cows, sheep.) (1, 1) solution, which is given. = Alcuin 5. 120. 20 for 20 at 3, 3/2, ½. (Oxen, sheep, geese.) (3, 1) solutions, (1, 1) given. = Bakhshali. 121. 12 for 12 at 2, ½, ¼. (Knights, boys, girls eating bread.) (2, 1) solutions, (1, 1) given. = Alcuin 47. 122. 12 for 12 at 2, 1, ½, ¼. (Knights, men, boys, girls dividing a bill?) (11, 4) solutions, (1, 1) given. = Munich 14684, XII. 123. 30 for 30 at 3, 2, ½. (Men women, children.) (3, 1) solutions, (1, 1) given. = Alcuin 33. 124. 100 for 100 at 3, 2, ½. (Men, women, children.) (7, 6) solutions, (1, 1) given. = Alcuin 34. 125. 90 for 90 at 3, 2, ½. (Men, women, children.) (7, 5) solutions, (1, 1) given. = Alcuin 33A. 126. 100 for 100 at 3, 1, 1/24. (Horses, oxen, sheep.) (2, 1) solutions, (1, 1) given. = Alcuin 38. 309. same as 81, done in three different ways.
Correspondence of Johannes Regiomontanus, 1463?-1465. Edited by Maximilian Curtze as: Der Briefwechsel Regiomontan's mit Giovanni Bianchini, Jacob von Speier und Christian Roder. Part II of: Urkunden zur Geschichte der Mathematik im Mittelalter und der Renaissance. AGM 12 (1902).
P. 262, letter to Bianchini, nd [presumably 1464]. 240 for 16047 at 97, 56, 3, specifically ruling out fractions, but not describing anything bought. Pp. 293 & 296, letter to von Speier, nd [apparently early 1465]. On p. 293, Regiomontanus invites his friend to a dinner of pheasants, partridges and wine (see Hermelink, op. cit. in 3.A). P. 296, same problem with (pheasants, partridges, doves). P. 300, letter from von Speier, 6 Apr 1465. Gives the only answer: 114, 87, 39.
Benedetto da Firenze. c1465. Pp. 151 152. Part of the text is lacking, but the problem must be 60 for 60 at 6, ½, ⅓. (Cows, calves, pigs.) This has (2, 2) solutions, (1, 1) is given.
Gottfried Wolack. 1468. Dresden MS C80, ff. 301'-303. This is a MS of lectures given at Erfurt, 1467 & 1468, Transcribed and discussed in: E. Wappler; Zur Geschichte der Mathematik im 15. Jahrhundert; Zeitschrift für Math. und Physik -- Hist.-Litt. Abteilung 45 (1900) 47-56. P. 51 has: 100 for 100 at 2, 1, ½. (Men, women, children.) There is an obscure calculation leading to dividing the people in the proportion 4 : 2 : 1, so there are 57 1/7 men! This may be getting confused with 7.G.2 -- Posthumous twins.
della Francesca. Trattato. c1480.
Ff. 13r-13v (58). Combining metals: 60 for 480 at 5, 7, 9. (16, 14) solutions, (1, 1) given: 10, 10, 40. = Lucca 1754, 46v, with different solution. F. 14v (60). Combining metals. = Lucca 1754, 48v. Ff. 14v-15r (60-61). Combining grain, stated as: 30 for 300 at 7, 11, 13, 15, 16, but he solves for prices 7, 9, 11, 13, 15, 16. (1491, 289) solutions, but he gives a non-integral solution: (60, 90, 10, 10, 10, 30)/7. F. 16v (63-64) = ff. 37v (98). Two types of cloth. 7 for 100 at 12, 15. Answer: (16, 5)/3. English in Jayawardene. F. 19r (67-68) = f. 44r (108). Three melons less a watermelon are worth 6, while 7 melons and 10 watermelons are worth 28, i.e. 3m - w = 6, 7m + 10w = 28. I include this as an early example of the use of a negative coefficient in such problems. Answer: (88, 42)/37. English in Jayawardene. Ff. 19v-20r (68-69). 7 geese, 6 hens and 8 partridges cost 240 but 3 geese, 5 hens and 10 partridges cost 480. Though not really a hundred fowls problem, one elimination step leads to one equation in two unknowns just as the hundred fowls does. (1, 1) solution, given. F. 20v (69-70). 100 for 100 at ½, ⅓, 1, 3. (Pearls, rubies, sapphires, balas-rubies.) This has (276, 226) solutions -- he gives (1, 1): 8, 51, 22, 19. English in Jayawardene. Cf Pacioli, Summa, 17.
Luca Pacioli. Aritmetica. c1480. ??NYS -- described in Sesiano. F. 238r.
10 for 10 at 2, 3. Answer: 20, -10. 10 for 10 at ½, ⅓. Answer: 40, -30.
Chuquet. 1484.
Triparty, part 1. Of apposition and remotion. English in FHM 88-90. Several examples treated as problems in numbers -- no mention of buying anything. Gives a method of finding one positive solution.
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