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).

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 a_i 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, ..., x_n and we know x + x_i = a_i and x + x₁ + ... + x_n = s. Then x  =  (a₁ + ... + a_n   s)/(n 1). (Iamblichus uses n unknowns.) Heath (HGM I 94 96) says Iamblichus continues and applies the Bloom to I (a₁, ..., a_n), with integral a_i, by letting x be the value of the purse and s = (a₁+1)...(a_n+1), which yields x + x_i  =  sa_i/(a_i+1). (We can take n-1 times the value of s to insure integer solutions.) For rational a_i, we let s be the LCM of the denominators of a_i/(a_i+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 + ... + a_n)/(n 1).

Aryabhata. 499. Chap. II, v. 29, pp. 71-72. (Clark edition: pp. 40 41.) III (a₁, ..., a_n). Gives T = (a₁+...+a_n)/(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, ..., x_n. See the discussion under Iamblichus.

IV-(a₁, ..., a_n). 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   x_i = c   d_ix_i. 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 b_i 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 b_i of the purse, I'd have a_i times the rest of you", with a₁, b₁; a₂, b₂ = 2, ½; 3, ⅔. Answer: 11, 13; 30

V. 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 a_i 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).