7. arithmetic & number theoretic recreations a. Fibonacci numbers
No. 17, p. 135 is the same for four numbers. He does III (22, 24, 27, 20)
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- T x i = c d i x i . These are variations of Type III problems or of the Present of Gems problem, section 7.P.4.
- V. 159, pp. 136 137. III (22, 23, 24, 27).
- V. 244, p. 157. I (2, 3). Answer: 3, 4; 5.
- V. 249. As in v. 242, with 2, ¼; 3, ⅓; 4, ½. Answer: 55, 71, 66; 876.
- 30: same as 29 with four men and purses (20, 30, 40, 50).
- Pp. 212 213 (S: 317-318). I (3, 4). Answer: 4, 5; 11.
- P. 216 (S: 322-323). II (2, 3, 4). Answer: 9, 16, 13; 23.
- Pp. 223 224 (S: 331-332). Three men purses: p, p+10, p+13, with A = (2, 3, 4).
No. 17, p. 135 is the same for four numbers. He does III (22, 24, 27, 20).No. 18 & 19 are like 3 and 4 men finding 3 and 4 purses -- see under 7.R.No. 20, pp. 136 137. This is Type I with a purse of 0, i.e. "I have ai times the rest of you". He does this as I (3, 4, a3) since a3 is determined by a1 and a2.Iamblichus. On Nicomachus's 'Introductio Arithmetica'. c325. Pp. 62 63, ??NYS. Partly given in SIHGM I 138 141. Describes the 'Bloom of Thymaridas' which has n+1 unknowns x, x1, ..., xn and we know x + xi = ai and x + x1 + ... + xn = s. Then x = (a1 + ... + an s)/(n 1). (Iamblichus uses n unknowns.) Heath (HGM I 94 96) says Iamblichus continues and applies the Bloom to I (a1, ..., an), with integral ai, by letting x be the value of the purse and s = (a1+1)...(an+1), which yields x + xi = sai/(ai+1). (We can take n-1 times the value of s to insure integer solutions.) For rational ai, we let s be the LCM of the denominators of ai/(ai+1). Iamblichus gives the problems I (2, 3, 4) and I (3/2, 4/3, 5/4). See Chuquet for an indeterminate version of the Bloom. This is closely related to problems like the following: x + y = a, y + z = b, z + x = c, i.e. III (b, c, a) = IV-(a, b, c). We set T = x + y + z and so T z = a, T y = b, T x = c. This is a case of the 'bloom' for n = 3, with x = T, x1 = z, a1 = a, etc., and s = 0. In general, this gives us x = T = (a1 + ... + an)/(n 1). Aryabhata. 499. Chap. II, v. 29, pp. 71-72. (Clark edition: pp. 40 41.) III (a1, ..., an). Gives T = (a1+...+an)/(n 1). Bakhshali MS. c7C. Kaye I 39-42, sections 78-79 and Datta, pp. 45 46 discuss two types of related systems for n amounts x1, x2, ..., xn. See the discussion under Iamblichus. IV-(a1, ..., an). When n = 3, this is equivalent to type III. Kaye notes that n is always odd and says the following occur: IV-(13, 14, 15) (Kaye III 166, f. 29r); IV (16, 17, 18, 19, 20) (Kaye III 166-167, ff. 29v & 27v); while the following are implied: IV (9, 5, 8); IV (70, 52, 66); IV-(1860, 1634, 1722); and possibly IV (36, 42, 48, 54, 60). Kaye's concordance (I 38-39) implies these examples should be in the same area of the text, but I can't find them in his Part III -- ??. Also III-(317, 347, 357, 362, 365) (Kaye I 40 (omitting the fourth equation); III 168 169, ff. 1v-2r).T xi = c dixi. These are variations of Type III problems or of the Present of Gems problem, section 7.P.4.Bhaskara I. 629. Commentary to Aryabhata, chap. II, v. 29. Sanskrit is on pp. 125-127; English version of the examples is on pp. 307-308. Ex. 1: III-(30, 36, 49, 50). Ex. 2: III-(28, 27, 26, 25, 24, 23, 21). Mahavira. 850. Chap. VI, v. 159, 233 250, pp. 136 137, 153 158. V. 159, pp. 136 137. III (22, 23, 24, 27).V. 236, p. 155. I (2, 3, 5). Answer: 1, 3, 5; 15.V. 239, p. 156. i-th says "If I had bi of the purse, I'd have 3 times the rest of you", with B = (1/6, 1/7, 1/9, 1/8, 1/10). Answer: 261, 921, 1416, 1801, 2109; 110880.V. 242, p. 157. i-th says "If I had bi of the purse, I'd have ai times the rest of you", with a1, b1; a2, b2 = 2, ½; 3, ⅔. Answer: 11, 13; 30V. 244, p. 157. I (2, 3). Answer: 3, 4; 5.V. 245, p. 157. I (8, 9, 10, 11). Answer: 103, 169, 223, 268; 5177.V. 248, pp. 157 158. As in v. 242, with four men and a1, b1; ... = 2, 1/5; 3, 1/4; 5, 1/2; 4, 1/3. Answer: 356, 585, 445, 624; 14760.V. 249. As in v. 242, with 2, ¼; 3, ⅓; 4, ½. Answer: 55, 71, 66; 876.al Karkhi. c1010. Sect. III, no. 24 25 & 29 30, pp. 95 & 98. 24: III (20, 30, 40). (= Diophantos I 16.)25: III (30, 45, 40, 35).29: Three men find purses (30, 40, 20). i-th says: "If I had the i th purse, I'd have as much as all of you."30: same as 29 with four men and purses (20, 30, 40, 50).Tabari. Miftāh al-mu‘āmalāt. c1075. P. 129, no. 46. ??NYS -- Hermelink, op. cit. in 3.A, says this is a problem with two persons. Tropfke 607 cites this with no details. Fibonacci. 1202. Pp. 212 228 (S: 317-337), Chap. XII, part 4: De inventione bursarum [On the finding of a purse]. Many problems, going up to five men, 4 purses and an example with a negative solution. Pp. 212 213 (S: 317-318). I (3, 4). Answer: 4, 5; 11.Pp. 213 214 (S: 318-319). I (2, 3, 4). Answer: 7, 17, 23; 73. (= Iamblichus's 1st.)Pp. 214 215 (S: 320). I (3, 4, 5, 6). Answer: 4, 67, 109, 139; 941.Pp. 215 216 (S: 320-322). I (5/2, 10/3, 17/4, 26/5, 37/6). Answer: 49154, 30826, 89478, 131962, 163630, 1088894. "... aut positio huius questionis indissolubilis est; aut primus homo debitum habebit ...." [... this posed problem will not be solvable unless the first man has a debit ...]. See Sesiano.P. 216 (S: 322-323). II (2, 3, 4). Answer: 9, 16, 13; 23.P. 217 (S: 323). II (5/2, 10/3, 17/4). He doesn't find the answer which is 284, 444, 381; 826.Pp. 218 220 (S: 325-327). II (2, 3, 4, 5). Answer: 33, 76, 65, 46; 119.Pp. 220 223 (S: 327-330). Four men find four purses of values:p1, p2, p3, p4 = p, p+3, p+7, p+13.i-th says "If I had the i-th purse, I'd have ai times the rest of you", with A = (a1, a2, a3, a4) = (2, 3, 4, 5).P. 223 (S: 330-331). Two men & purses of values: p, p+13, with A = (2, 3).Pp. 223 224 (S: 331-332). Three men & purses: p, p+10, p+13, with A = (2, 3, 4).Download 1.54 Mb. Share with your friends:
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