7. arithmetic & number theoretic recreations a. Fibonacci numbers

No. 15. pp. 134 135. "To find two numbers such that each after receiving from the other may bear to the remainder a given ratio." Does (30, 2; 50, 3).

No. 18, pp. 135 136. "To find three numbers such that the sum of any pair exceeds the third by a given number." E.g. "If I had 20 more, I'd have as much as you two." Does with values 20, 30, 40. This is like finding several purses -- see 7.R.1.

No. 19, pp. 136. Same as no. 18, with 4 people. Does with values 20, 30, 40, 50.

Metrodorus. c510. Art. 145 146, p. 105. (10, 3; 10, 5); (2, 2; 2, 4). Earliest example with non-integral answers.

Alcuin. 9C. Problem 16: Propositio de duobus homines boves ducentibus. (2, 1; 2, 2), but the second statement in the problem is interpreted as happening after the first is actually carried out. If a problem with parameters (a, b; c, d) is interpreted this way, it is the same as our usual problem with parameters (a, b; c-a, d).

Mahavira. 850. Chap. VI, v. 251 258, pp. 158 159.

253. I (9, 2; 10, 3; 11, 5).

256. I (25, 3; 23, 5; 22, 7).

al Karkhi. c1010. Sect. III, no. 5, p. 90. II (1, 2; 2, 3; 3, 4; 4, 5).

Tabari. Miftāh al-mu‘āmalāt. c1075. P. 145f., no. 7. ??NYS -- cited by Tropfke 611.

Bhaskara II. Bijaganita. 1150. Chap. IV, v. 106. In Colebrooke, p. 191. (100, 2; 10, 6).

Fibonacci. 1202. Pp. 190 203 (S: 289-305). Numerous versions, getting up to five people, some inconsistent examples and types where the second clause is "I'd have b₁ times you plus b₂ more."

Pp. 190 (S: 289-290). (1, 1; 1, 10).

Pp. 190 191 (S: 290-292). (7, 5; 5, 7) and general rules.

Pp. 191-192 (S: 292). (6, 5¼; 4, 7⅔).

P. 192 (S: 292-293). (7, 5, 1; 5, 7, 1) -- denoting the extended type where the first says "If I had 7 from you, I'd have 5 times you plus 1 more". I.e. the first equation is x + 7  =  5 (y   7) + 1.

Sesiano analyses the extended type on pp. 192-198 in detail.

Pp. 197 198 (S: 298-300). (7, 5,  1; 5, 7,  3).

Pp. 198 199 (S: 300-301). I (7, 5; 9, 7; 11, 7), but he solves I (7, 5; 9, 6; 11, 7).

Pp. 199 200 (S: 301-302). I (7, 5, 1; 9, 7, 1; 11, 7, 1), but he solves as though the middle 7 is a 6.

Pp. 200 201 (S: 302-303). Types giving x + y + 7 = 5 (z   7) etc.

P. 201 (S: 303-304). De eodem inter quattuor homines questio insolubilis [On the same with four men, an unsolvable problem]. This gives equations like w + x + 7 = 3 (y + z - 7) with coefficients 7, 3; 8, 4; 9, 5; 11, 6. This is indeed inconsistent.

Pp. 202 203 (S: 304-305). I (7, 2; 8, 3; 9, 4; 10, 5; 11, 7), but he solves with the last 7 as a 6.

Pp. 325 326 (S: 455-456). Problem of pp. 190 191, (7, 5; 5, 7), done by false position.

Pp. 332 333 (S: 463-465). I (7, 4; 9, 5; 11, 6).

Pp. 344 346 (S: 477-480). I (7, 3; 9, 4; 11, 5) done in two ways.

Lucca 1754. c1330. Ff. 29r 30v, pp. 68 70. (12, 2; 17, 3). II (15, 2; 18, 3; 21, 5).

Munich 14684. 14C. Prob. XV, p. 80. (1, 1; 2, 2) and (n, 1; n, 2).

Folkerts. Aufgabensammlungen. 13-15C. Many sources. Almost all have (a, 1; a, 2), with the objects being exchanged being, gold, animals, coins, nuts, often noting that the answer is a times the answer for a = 1. Examples of (a, 1; a, d) with d = 3, 5, 9. 17. An example of (a, ½; a, d). Folkerts cites Metrodorus, Alcuin, Fibonacci, AR, etc.

Three sources of the following. Ask a person to put the same amount of money into each of her hands. Tell her to transfer n coins from the right hand to the left. Now transfer enough from the left to double what is in the right hand. This leaves 2n in the left hand.

Giovanni di Bartolo. Op. cit. in 7.H. c1400. He gives complex examples in probs. 10 14, 54, 56, 57 on pp. 18 27, 101 107. E.g. prob. 10, pp. 18 21. "If I had the square root of your money, I'd have 3 times you." "And if I had the square root of your money, I'd have 4 times you."

Provençale Arithmétique. c1430. Op. cit. in 7.E.

F. 114r, p. 60. (1, 1; 1, 2).

F. 114v, p, 61. (1, 2; 1, 4). This has non-integral answers.

F. 115r, p. 61. II-(1, 1; 1, 2; 1, 3). Non-integral answers.

Pseudo-dell'Abbaco. c1440.

Prob. 69, pp. 63 65 with simple plate on p. 64. (8, 2; 10, 3).

Prob. 126, pp. 100 102 with simple plate on p. 101. (3, 2; 5, 3).

AR. c1450. Prob. 138 147, 220, 334 336. Pp. 70 71, 102, 146 147, 169 171, 218.

138. Regula augmentationis: (1, 2; 1, 3); (10, 2; 10, 3).

139 147. (3, 4; 4, 3); (3, 3; 4, 4); (3, 2; 5, 3); (5, 2; 3, 3); (1, 1; 1, 3); (15, 2;  15, 3); (3, 3; 3, 5) -- erroneous answer; (1, 10; 1, 20); (1, 1; 1, 2).

220. same as 144.

334. I (3, 2; 3, 4; 3, 10; 3, 100) phrased as "If you each give me 1, ..."

335. I (3, 2; 4, 4; 6, 7; 5, 9).

336. I (2, 2; 2, 4; 2, 10).

He then says that I (1, 2; 1, 3; 1, 5) and I (1, 2; 1, 3; 1, 4) are impossible. However, I find solutions in both cases, though each person has  1 which may be why AR is unhappy. Vogel finds the same solutions that I do, but doubled because he reads the problems as I (2, 2; 2, 3; 2, 5) as in 334. I don't read them that way, but the text and numerical layout are a bit inconsistent.

Benedetto da Firenze. c1465. Pp. 153. (60, 6; 50, 13); II (10, 2; 19, 4; 15, 9); I (16, 2;  30, 8;  26, 7/2).

Muscarello. 1478. Ff. 81v-82r, p. 198. (2, 2; 2, 1).

della Francesca. Trattato. c1480.

Ff. 43r (106-107). The amount requested by the second two is a fraction of what the others have, so this is a mixed version of the problem leading to x + 6 = 2(y+z 6); y + 2(x+z)/3 = 3(x+z)/3; z + 3(x+y)/4 = 4(x+y)/4. Answer:  (198, 90, 72)/7.

Ff. 121r-121v (258-259). First says "If you give me that part of yours which is as 5 is to mine, then I will have 4 times you." I.e. x + 5y/x = 4(y-5y/x). The second makes a similar statement. The solution is obtained by a kind of false position, but I don't follow it. There is an arithmetic error in the last line.