7. arithmetic & number theoretic recreations a. Fibonacci numbers

No. 55, pp. 72 75. Buying a business worth 20, I-(4/9, 12/35). Answer: (10500, 12420)/801

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No. 55, pp. 72 75. Buying a business worth 20, I-(4/9, 12/35). Answer: (10500, 12420)/801.

No. 56, pp. 74 77. Same, I-(12/35, 13/42), with h = 72. Answer: (69552, 73080)/1314.

No. 57, pp. 78 81. Same, I-(71/105, 37/60), with h = 50. Answer: (102000, 120750)/3673.

No. 58, pp. 80 83. A gives B 7/12 of what A has, then B returns 9/20 of what he has, then both have 12. Answer: (288, 1032)/55.

No. 72, pp. 94 97. Same with constants 1/7, 1/4, and both finish with 36. Answer: 28, 44.

No. 114, pp. 130 131. (= Fibonacci, p. 229.)

Gherardi. Libro di ragioni. 1328.

P. 42. Same as Fibonacci, pp. 235 236, with h_i = 10, 12, A = (⅓, ¼). Answer: (72, 114)/11.

Pp. 45 46. Chopera. II (1/2, 1/3, 1/4, 1/5). Assumes house is worth 60. Answer is 60/119 times de Nemore's II 25.

Pp. 46 47. I (½, ⅓, ¼). Answer: 5, 11, 13; 17.

Pp. 59 60. Three men and three horses: I (1/3, 1/4, 1/5) with h_i = 40, 47, 55. Answer: (702, 1602, 2289)/50.

Lucca 1754. c1330.

F. 58r, p. 131. Buying a horse. II (½, ⅓, ¼). Answer: 16, 18, 21; 25.

F. 58r, pp. 131 132. Buying a horse. I (½, ⅓, ¼). Answer: 5, 11, 13; 17. Cf Gherardi, pp. 46-47.

F. 58v, p.132. Two men. First says: "If you give me ⅓ of your money, then I can buy 20 horses." Second says: "If you give me ¼ of your money, then I can buy 21 horses." Answer: 156, 192 with horses worth 11 each.

Ff. 61r 61v, p. 141. Three men and a friend. i-th says: "If I had a_i of our friend's money, I could buy the horse", with (a_i) = (½, ⅓, ¼). This has a two dimensional solution space. He gives only 1/2, 5/2, 7/2 with friend having 12 and horse worth 13/2.

Munich 14684. 14C. Prob. XVII, p. 80. I (½, ⅓, ¼). Answer: 10, 22, 26; 34. Cf Gherardi, pp. 46-47.

Folkerts. Aufgabensammlungen. 13-15C. He calls it Sperberkauf (sparrow hawk purchase). 11 sources for I (½, ⅓, ¼) = Munich 14684 = Gherardi, pp. 46-47, none of which give a derivation. It's not clear if h = 34 is given in some cases. Numerous other citations.

Provençale Arithmétique. c1430. Op. cit. in 7.E. F. 100r-101v, pp. 49-53.

Three men buy a horse: I (½, ⅓, ¼). Answer: 5, 11, 13; 17. Cf Gherardi, pp. 46-47.

Four men buy a horse: I (1/2, 1/3, 1/4, 1/5) (= Fibonacci, 245-248). The answer here takes C = 60, which gives: (5, 95, 125, 140; 185)/3

Five men buy a piece of cloth: I (1/2, 1/3, 1/4, 1/5, 1/6). Answer: ( 43, 77, 117, 137, 149; 197)/4. Sesiano, loc. cit. in 7.E, asserts "Tel est le plus ancien exemple de l'acceptation d'une solution négative dans un texte mathématique, ...." The text says simply "restan 10 et ¾ mens de non res." Sesiano, loc. cit. at beginning of 7.R, gives this in English: "... the first work in which a negative result is admitted without any restriction, .... No interpretation whatsoever is given of the negative result. The only hint at its peculiarity is the -- exceptional -- verification of the results: ...."

AR. c1450. Prob. 156, 157, 171 181, 186, 221, 223, 224, 338 341. Pp. 74 75, 82 85, 87, 102 103, 149 150, 171 173, 218 219.

156: Regula posicionum: II (1/3, 1/4, 1/5) with horse worth 100. Makes two steps by false position and then gives up and goes to the next section.

157: II (1/3, 1/4, 1/5). Answer: 45, 48, 52; 61. = Fibonacci, p. 229.

171: I (⅓, ¼). Answer: 8, 9, 11. Cf Al-Karkhi I -42. Vogel says this occurs in Ibn al Haitham, ??NYS.

172 175: same as 171, with various values of horse: 31, 100, 81, 1.

176: I (⅓, ⅔) with horse worth 30. Cf Chiu Chang Suan Ching.

177: 173 redone.

178: Pferd mit 3 an namen (Horse with 3 [men] without price). I (½, ⅓, ¼). Answer: 10, 22, 26; 34. Cf Gherardi, pp. 46-47.

179: same heading, II (½, ⅓, ¼). Answer: 16, 18, 21; 25. Cf Lucca 1754 58r.

180: Pferd mit 3 [und] mit namen (horse with 3 [men and] price). II (½, ⅔, ¾) with horse worth 100. Answer: (200, 200, 150)/3.

181: 3 men buy a fish, leading to x + ¼ (y + z) = ½y + ¼ (x + z) = ⅓z + ¼ (x + y) = cost. Answer: 1, 3, 9; 4 and its multiples.

186: 3 men want to buy horses worth 10, 20, 30, as in Fibonacci, pp. 235 240, with A = (½, ⅓, ¼). Answer: (24, 52, 144)/5.

221: same as 173.

223: II (½, ⅓, ¼) with horse worth 51. See 179.

224: same as 178, with answer: 5, 11, 13; 17.

338: same as 157.

339: same as 179, with comment that multiplying the solution is again a solution.

340: same as 178.

341: I (1/3, 1/4, 1/5). Answer: (65, 85, 95; 125)/2. Cf Diophantos 24.

Dresden MS C80, 15C, has some of these problems. ??NYR -- mentioned in BR, p. 157.

Benedetto da Firenze. c1465.

Pp. 155 157.

I-(½, ⅓) with h_i = 60, 80.

I-(½, ⅓, ¼) with h_i = 60, 50, 163/5.

II-(½, ⅓, ¼) with h_i = 60, 50, 40.

Chap. 20: "... uomini che ànno d. et vogliono chompare chavagli," pp. 168 171.

I (1/6, 1/7). Answer: 35, 36; 41.

II (1/4, 1/6, 1/8). Answer: 152, 164, 174; 193. Illustration on p. 169.

I (½, ⅓, ¼). Answer: 5, 11, 13; 17. Cf Gherardi 46-47.

Muscarello. 1478. Ff 56v-58r, pp. 158-161. Four men buying a house worth 100, II (2/3, 5/8, 4/5, 7/10). Answer should be: (1250, 1575, 1160, 1425)/23, but there are two copying errors in the MS. The second answer is given as 66 11/23 instead of 68 11/23 and the fourth answer is given as 61 12/23 instead of 61 22/23.

della Francesca. Trattato. c1480.

F. 16r (62-63). I-(½, ⅓) with horse worth 35. Answer: 21, 28. English in Jayawardene.

F. 17v (64-65). II-(1/3, 1/4, 1/5) with horse worth 30, Answer: (1350, 1440, 1560)/61. English in Jayawardene. Cf Fibonacci, p. 229; AR 156, 157.

Ff. 17v-18r (65-66). I-(½, ⅓, ¼) with horse worth 30. Answer: (450, 990, 1170)/51. He doesn't notice that the fractions in the answers can be reduced. Cf Gherardi 46-47; Provençale Arithmétique.