12 for 12 at 2, 1, ½. = Prov. Arith. no. 1, which has (5, 3) solutions. 12 for 60 at 8, 5, 3. (3, 2) solutions. 12 for 36 at 4, 3, 2. (7, 5) solutions. Two further examples are in the margin and FHM just gives the formulae and one solution. 12 for 35 at 6, 3, 1 {FHM has 6, 1, 1 but this doesn't agree with their answer}. (2, 2) solutions. 20 for 20 at 6, 2, ½. (1, 1) solutions Chuquet says: "This rule cannot be extended for the discovery of four or more numbers. Also one should know that all such calculations have several answers, as many as you want, as appears [later in] this book. Wherefore apposition and remotion is a science which has little to recommend it."
Appendice.
Prob. 34. 15 for 160 at (9, 13). This is determinate and is only included as a lead-in to the next problem. Prob. 35. English in FHM 206-207. 15 for 160 at (11, 13). (Two kinds of cloth.) Answer: 35/2, -5/2. Chuquet gives an interpretation of this. See Sesiano, op. cit. in 7.R. FHM says the "explanation fails to be entirely clear or convincing". Prob. 40. English in FHM 208. Mixing of two kinds of waxes giving 100 for 1100 at (9, 14). Prob. 83. English in FHM 213-214. Reckoning which is done by the rule of apposition and remotion. 30 for 30 at 4, 2, ½. Same as Abbot Albert, p. 334. (2, 1) solutions. He gives solutions 3, 3, 24 and 4, ⅔, 25⅓ and says one can have as many as one wants. This holds because the products are cloth! Prob. 84. Mentioned in passing on FHM 214. 20 for 20 x 20 at 30, 25, 16, 18. (10, 5) solutions -- he gives (1, 1).
Borghi. Arithmetica. 1484. Ff. 93v-101v. Several problems of alligation, getting up to mixing five grains of values 44, 48, 52, 60, 66 per measure to mix to produce a product of value 50.
Johann Widman. Op. cit. in 7.G.1. 1489. ??NYS. Glaisher, pp. 14 & 121, gives: F. 109v: mix wines worth 20, 15, 10, 8 to make one worth 12. Gives one solution: 6, 6, 11, 11.
Pacioli. Summa. 1494.
F. 105r, prob. 14. 3 hens, 4 partridges and 5 geese cost 72; while 2, 5, 7 cost 94⅔. That is: 3x + 4y + 5z = 72, 2x + 5y + 7z = 94⅔. He gives one solution: (4, 16, 28)/3 with no indication that the problem is indeterminate. ⅓, 9, 7 is an easier solution. There is no integral solution and the number of rational solutions is infinite! Cf della Francesca 19v. Ff. 105r-105v, prob. 17. 100 for 100 at ½, ⅓, 1, 3. (Sheep, goats, pigs, asses.) This has (276, 226) solutions -- he gives (1, 1): 8, 51, 22, 19. = della Francesca 20v, with different objects, but same solution. F. 105v, prob. 18. 20 for 20 at 4, ½, ¼. (Men women, children) eating at a tavern. This has (2, 1) solutions -- he gives (1, 1). (See H&S 93.)
Riese. Rechnung. 1522. 1544 ed. -- pp. 104 106; 1574 ed. -- pp. 70r 71v. The 1544 ed. calls this section 'Regula cecis oder Virginum'; the 1574 ed. calls it 'Zech rechnen'. There is first a simple problem with only two types, hence determinate.
20 for 20 at 3, 2, ½. (Men, women, girls drinking.) (2, 1) solutions, (1, 1) given. = Alcuin 32. 100 for 100 at 4, 3/2, ½, ¼. (Oxen, pigs, calves, goats.) (265, 222) solutions, (1, 1) given: 12, 20, 20, 48. The 1574 ed. has a nice woodcut illustration.
Tonstall. De Arte Supputandi. 1522. P. 240. Repeats Pacioli's prob. 14, except there is a misprint -- in the second case, he has 3, 5, 7 costing 94⅔.
Riese. Die Coss. 1524. No. 67, p. 49. 100 for 460 at 3, 5. (Coins.)
H&S 93 says a tavern version is in Rudolff (1526?).
Apianus. Kauffmanss Rechnung. 1527.
Ff. H.viii.r - J.ii.v is Regula virginum. He describes how to eliminate the least valuable variable and then gives a feeble attempt at describing how to find a solution from the result. [No. 1.] 26 for 88 at 6, 4, 2. (Men women, girls.) (10,8) solutions. He gives (3,3) solutions and says more (all?) can be found. [No. 2.] 20 for 20 at 2, 1, 1/2, 1/4. (Men women, girls, children.) (25,14) solutions. He gives (3,3). [No. 3.] 100 for 200 at 4, 3, 5/2, 1. (Nutmeg, cinnamon, cloves, saffron & pepper.) (289, 256) solutions. He gives (1,1). [No. 4.] 300 for 2000 at 24, 12, 8, 4. (Ranks of soldiers.) (2081, 1921) solutions. He gives (1,1). Ff. J.vvi.r - K.viii.v is Regula Alligationis Ff. K.viii.v - L.ii.r is Munzschlagen. These two sections deal with mixing of wine, spices, metals, etc., getting up to seven types.
Sesiano cites a 16C MS Ambros. P114 sup which gives 40 for 100 at 1/5, 1/10. Answer: 960, -920.
Cardan. Practica Arithmetice. 1539.
Chap. 47, f. L.iiii.v (p. 71). End mentions Pacioli prob. 17. Chap. 66, section 35, f. DD.v.r (pp. 145-146). 100 for 100 at 3, 2, ½. (Pigs, asses, cows.) (1, 1) solution given. = Alcuin 34. Chap. 66, section 67, ff. FF.ii.v - FF.iii.v (pp. 155-156). (67 is not printed in the Opera Omnia.) 100 for 100 at 3, ½, ⅓, 1/11. (Turtledoves, thrushes, crested larks, sparrows.) (18, 15) solutions, (1, 1) given.
Tartaglia. General Trattato. 1556. Book 16, art. 117 129, pp. 254r 255v & Book 17, art. 25, 26, 43, 44, pp. 272v & 277r-277v. 18 versions. The objects being bought are mostly not in the Italian dictionaries that I have consulted. They are apparently 16C Italian, probably Venetian dialect. Several Italian or Italian-speaking friends have helped to determine these -- my thanks to Jennifer Manco, Ann Maury, Ann Sassoon and especially Maria Grazia Enardu and a student of hers.
16-117. 60 for 60 at 4, 2, ½. (Thrushes, larks, redstarts.) (3, 2) solutions, he gives (1, 1). 16-118. 20 for 20 at 3, 2, ½. (Partridges, pigeons, quails.) (2, 1) solutions, he gives (1, 1). = Alcuin 32. He discusses fractional solutions and gives one, but says it is not really acceptable. 16-119. 20 for 240 at 18, 10, 3. (Sorghum, bran, grape seeds.) (2, 0) solutions. He finds 5, 15, 0, then rejects it and finds a fractional solution but refers back to the previous discussion.
Dostları ilə paylaş: |
