7. arithmetic & number theoretic recreations a. Fibonacci numbers
Download 1.54 Mb.
|
- Navigate this page:
- 57. (7, 2; 9, 6). 58. (20, 2; 30, 3).
- 77. II-(-9, 1/6; -11, 1/7; -7, 1/5) - i.e. the first equation is y + z + 9 = 6 (x - 9).
- 79. This has equations like v + w + x + 7 = 2 (y + z - 7) with coefficients (7, 2; 8, 3; 9, 4; 10, 5; 11, 6).
- Ff. 189r-189v, prob. 12. x + 12 = 2 (y - 12), y + x(12/y) = 3 {x x(12/y)}
- F. 193v, prob. 34. x + 6 = 2 (y + z - 6), y + ⅔ (z + x) = 3 {⅓ (z+x)}, z + ¾ (x + y) = 4 {¼ (z + x)}. Answer: (198, 90, 72)/7.
- Prob. 1, f. 100r. (10, 1; 20, 2).
- Prob. 9, f. 102r. Same as prob. 7, but the extra condition is replaced by x + y = 30.
Ff. 121v-122r (259-260). Two similar problems with a mixture of ordinary statements giving a fixed amount or a fixed part of the others money and statements as in the previous. Again, I don't follow the solutions and the second leads to a quadratic.Chuquet. 1484. Prob. 57, 58, 59, 60. Prob. 61 79 extend in various ways. English of 69, 70 78 in FHM 210-212. 78 is indeterminate. 57. (7, 2; 9, 6).58. (20, 2; 30, 3).59. (1, 1; 1, 2).60. (2, 1; 3, 4).61. II-(3, 2; 4, 3; 5, 5).62. II-(2, 2; 2, 3; 2, 4).63. I-(7, 5; 9, 6; 11, 7).64-76 lead to equations like x + 7 = 5 (y - 7) + 1 or 5 (y + z - 7) + 1.77. II-(-9, 1/6; -11, 1/7; -7, 1/5) - i.e. the first equation is y + z + 9 = 6 (x - 9).78. This has equations like w + x + 100 = 3 (y + z - 100) with coefficients (3, 100; 4, 106; 5, 145; 6, 170), which is indeterminate. "Thus it appears that such problems have a necessary answer for two by two, but for one by one they have whatever answer one desires."79. This has equations like v + w + x + 7 = 2 (y + z - 7) with coefficients (7, 2; 8, 3; 9, 4; 10, 5; 11, 6).Calandri. Arimethrica. 1491. F. 66r. (20, 2; 30, 3). This has the unusual feature that x = y and I do not recall any other such example. The condition for x = y is more complex than one might expect: a (b+1)/(b-1) = c (d+1)/(d-1). Pacioli. Summa. 1494. He has numerous problems, sometimes mixing amounts and parts and sometimes mixing this topic with 7.R.1 and 7.R.2, often saying "that part of yours that 12 is to mine", i.e. y(12/x) {cf. della Francesca and 7.R.1}, and he often continues into problems where one gives the square root of what one has or says something about the square of an amount. F. 105v, prob. 19. Two men find two purses of values p+10, p, giving equations: x + p+10 + 10 = 4 (y - 10), y + p + 20 = 5 (x - 20). He assumes the purses are worth 100 in total, so p = 45, p+10 = 55. Answer: (765, 690)/19.Ff. 189r-189v, prob. 12. x + 12 = 2 (y - 12), y + x(12/y) = 3 {x x(12/y)}F. 191r-191v, prob. 24 & 25. x + y(20/x) = y - y(20/x) + 28, y + x(30/y) = x x(30/y) + 70. Answer: 100, 120. Prob. 25 is an alternative way to solve the problem.F. 192v, prob. 29. x + ⅓ (y + z) = 96, y + 60 = 2 (z + x - 60) - 4, z + ¼ (x + y) + 5 = 3 {¾ (x + y)} - 5. Answer: -79, 236, 289. His algebra leads to 79 + x = 0.F. 193v, prob. 34. x + 6 = 2 (y + z - 6), y + ⅔ (z + x) = 3 {⅓ (z+x)}, z + ¾ (x + y) = 4 {¼ (z + x)}. Answer: (198, 90, 72)/7.Calandri, Raccolta. c1495. Prob. 23, pp. 22 23. (20, 2; 30, 3).Prob. 44, pp. 40 41. Three people -- first two as (12, 2; 20, 3), third says "If I had 24 from you two, I'd have 3 times you plus the square root of what you have."Hans Sachs (attrib.). Useful Table-talk, or Something for all; that is the Happy Thoughts, good and bad, expelling Melancholy and cheering Spirits, of Hilarius Wish-wash, Master-tiler at Kielenhausen. No publisher, place or cover, 1517, ??NYS -- discussed and quoted in: Sabine Baring-Gould; Strange Survivals Some Chapters in the History of Man; (1892), 3rd ed., Methuen, 1905, pp. 220-223. [Not in Santi.] Baring-Gould, p. 221 has (1, 2; 1, 1). Ghaligai. Practica D'Arithmetica. 1521. He has a series of problems of this type, of increasing complexity, all involving men and money. I omit the more complex cases. He also uses parts as in Pacioli. Prob. 1, f. 100r. (10, 1; 20, 2).Prob. 2, f. 100r. (20, 2; 30, 3).Prob. 3, f. 100v. x + ¼y = y - ¼y, y + ½x = 4 (x - ½x) + 2.Prob. 5, f. 101r. x + 10 = y - 10, y + x(20/y) = 3 {x - x(20/y)}.Prob. 6, f. 101r. x + 12 = 2 (y - 12), y + x(12/y) = 3{x - x(12/y)}. = Pacioli 12.Prob. 7, f. 101v. x + y(6/x) = 21, y + x(3/y) = 20, given that y(6/x) + x(3/y) = 11. [Without the extra condition, this gives a fourth order equation with solutions x = 0, 12, 15 ± 6; y = 0, 18, 11 -/+ 36, though 0, 0 is indeterminate in the original equations.Prob. 9, f. 102r. Same as prob. 7, but the extra condition is replaced by x + y = 30.Prob. 10, f. 102r-102v. x + ry = 2 (y - ry), y + rx = 5 (x - rx), where r is an unspecified ratio. This seems to be a unique version of this problem and both forms I and II lead to interesting solutions -- r is determined by the coefficients 2, 5. Cardan has a related version, but he gives a value of r which is inconsistent.For the 'all others' (type I) version, we let T = Σ xi. Then the equations are: xi + r(T - xi) = ai(1-r)(T - xi). Since xi + r(T - xi) + (1-r)(T - xi) = T, we see that xi + r(T - xi) = aiT/(1+ai), so T = (1+ai)[xi + r(T xi)]/ai = (1+ai)(1 r)(T-xi) or T - xi = T/(1+ai)(1-r), assuming 1-r 0. Adding these last equations gives (n 1)T = [T/(1 r)] Σ 1/(1+ai). Assuming T 0 gives us 1 r = [1/(n 1)] Σ 1/(1+ai). Hence r is determined by the ai's, or else T = 0 and/or 1-r = 0. In fact T = 0 holds if and only if 1-r = 0 or all xi = 0. When 1-r = 0, the xi are arbitrary. In either of these degenerate cases, the ai are arbitrary.For the 'next one' (type II) version, the equations are: xi + rxi+1 = ai(1 r)xi+1, or xi = [(1-r)(1+ai) - 1] xi+1. Multiplying these together, we find that the product of the factors must be 1 and this gives an n-th order polynomial for 1-r. Set P(x) = Π [(1+ai)x - 1]. If we assume all 1+ai > 0, then this has n positive roots and hence P(x) = 1 occurs for some x = 1-r greater than the largest term 1/(1+ai). If we further assume the ai are not too small, namely that Π ai > 1, then this product is P(1) and we hence know there is a point with P(x) = 1 for some positive x less than 1, so the corresponding r = 1-x is also between 0 and 1. There may well be other suitable roots. If some 1+ai < 0, the situation is more complex and there need not be any positive roots. Even if all ai are positive, P(x) = 1 may only occur for x > 1, and hence r < 0. When n is even, the constant terms in the equation cancel and one can factor out 1-r (since 1-r = 0 leads to r = 1, and an easy solution or inconsistency). E.g., for n = 2, we get 1 r = 1/(1+a1) + 1/(1+a2).Riese. Rechnung. 1522. 1544 ed. -- p. 93; 1574 ed. -- pp. 62v 63r. (1, 1; 1, 3). Riese. Die Coss. 1524. No. 25 30, p. 44 & No. 63 64, p. 49. Download 1.54 Mb. Share with your friends:
|