7. arithmetic & number theoretic recreations a. Fibonacci numbers

F. 21r (70-71). II-(1/3, 1/4, 1/5) with h_i = 12, 15, 18. Answer:  (510, 666, 996)/61.

F. 21v (71-72). I-(½, ⅔, ¾) with h = 12. Answer: (60, 36, 12)/7.

Ff. 22r-22v (72-73). Three men buying a horse. Leads to equations:

x + y + z/2 = x + y/3 + z = x/4 + y + z = 30. Answer: (240, 270, 360)/23

F. 36v (95-96). I-(⅓, ¼) with jewel worth 30. Answer: (240, 270)/11. Cf al-Karkhi III-26.

F. 36v-37r (96). II-(½, ⅓, ¼) with horse worth 35. Answer: (112, 126, 147)/5. Cf Lucca 1754 58r.

Ff. 37v-38r (98-99) = f. 21r.

F. 39r (100-101). I (½, ⅓, ¼) with horse worth 30. = ff. 17v-18r. Answer:  (150, 330, 390)/17. Cf Gherardi 46-47.

F. 39v (101-102). Three men buying a horse. Leads to equations:

x + y + z/3 = x + y/4 + z = x/5 + y + z = h. Answer: 30, 32, 36; 74. He doesn't notice that this can be divided through by two.

F. 40v (102-103). 3 men want to buy a horse and have a friend with p in his purse, giving: x + y + p/3  =  h, x + z + p/4  =  h, y + z + p/5  =  h. Answer:  12, 15, 20; 47 with 60 in the purse. Cf Lucca 1754 61r-61v.

F. 41r (103-104). x + (y + z)/3 + 1  =  14, y + (z + x)/4   2  =  17,

z + (x + y)/5 + 3  =  19. Equivalent to II-(1/3, 1/4, 1/5) with h_i = 13, 19, 16. Answer:  197/50, 2244/150, 1833/150.

F. 42r (105). x + y/2 + z/3 = 12, x/3 + y + z/4 = 14, x/4 + y/5 + z = 18. Answer:  (456, 2040, 3384)/217.

Ff. 44r-44v (108-109). x + y + z/2 + 1 = 20, x + y/3 + z - 2 = 20, x/4 + y + z + 3 = 20. Answer: (228, 84, 250)/23.

Chuquet. 1484. Triparty, part 1. Sesiano cites p. 641, ??NYS

I-(½, ⅓, ¼) with object worth 30. English in FHM 79. Answer:  (150, 324, 390)/17. Cf Provençale Arithmétique. 324/17 is given as 19 1/17 in FHM -- the correct value is 330/17 = 19 7/17, so the 7 has been misread as a 1 at some stage.

FHM 79-80 then simply states and discusses the next two, which are discussed by Sesiano.

I-(2/3, 3/4, 4/5, 5/6) with horse worth 40. Answer: 24, 16, 8, 0.

I-(1/2, 2/3, 3/4, 4/5, 5/6) with horse worth 40. Answer:  0, 20, 10, 0,  10. Chuquet then explains how to deal with negatives and zero.

FHM 81 gives the English of a version with equations x + y + ½z  =  ⅓x + y + z  =  x + ¼y + z = 20. Answer:  (180, 160, 240)/23.

FHM 81-83 then gives the next two by formulae with solutions, then then gives the English of the next.

II-(½, ⅓, ¼) with object worth 20. Answer: (64, 72, 84)/5. Cf Lucca 1754 58r.

II-(½, ⅔, ¾) with object worth 30. Answer:  (600, 540, 495, 500)/29. Cf AR 180.

II-(1/3, 1/4, 1/5) with object worth 20. Answer:  (900, 960, 1040)/61. Cf Fibonacci 229.

Borghi. Arithmetica. 1484. Ff. 113v-116v (1509: ff. 95v-98r). I (½, ⅓, ¼) with h = 20. Answer: (100, 220, 260)/17.

Calandri. Arimethrica. 1491. F. 667v. Two men buy a lamprey. I-(1/3, 1/5). Takes h = 60 rather arbitrarily and gets answer: (300, 360)/7.

Francesco Pellos. Compendion de lo Abaco. Turin, 1492. ??NYS -- see Rara 50-52 & both Sesiano papers mentioned at Provençale Arithmétique, c1430, above. Sesiano says ff. 64v-65r translates the three problems in the Provençale Arithmétique and is the first printed problem with a negative solution. He says it may have been composed c1460. Smith doesn't mention either of these points.

Pacioli. Summa. 1494.

Ff. 105v-106r, prob. 23. I-(½, ⅓, ¼) with h = 20. Answer: (100, 220, 260)/17. Cf Gherardi 36-37.

F. 190v, prob. 21. Two men find a purse, but leads to: x + ⅓y  =  p, y + ¼x  =  p, so this is I-(⅓, ¼). He assumes x + y + p = 100 and gets answer: (200, 225; 275)/7. Cf al-Karkhi I-24.

Ff. 191v-192r, prob. 26. x + ½ (y + z)  =  90, y + ⅓ (z + x)  =  84, z + ¼ (x + y)  =  81. I.e. I-(½, ⅓, ¼) with h_i = 90, 84, 81.

Answer: (576, 900, 1008)/17.

F. 192r, prob. 27. Same as prob. 26 with all values equal to 50. I.e. I-(½, ⅓, ¼) with h = 50. Answer:  (250, 550, 650)/17. Cf Gherardi 36-37.

F. 192v, prob. 31. I-(½, ⅓, ¼). Notes that the common value, h, can be set, as in prob. 23 (cf. 7.R.1) to 50, but when fractions appear, he converts to answer:  (5, 11, 13; 17). Cf Gherardi 36-37.

F. 193r, prob. 35. x + (y + z)/3 + 1  =  14, y + (z + x)/4   2  =  17, z + (x + y)/5 + 3  =  19. Equivalent to II-(1/3, 1/4, 1/5) with h_i = 13, 19, 16. Answer:  (197, 748, 611)/50. Cf della Francesca 41r.

F. 193v, prob. 37. x + y/2 + z/3  =  12, y + z/3 + x/4  =  15, z + x/4 + y/5  =  20. Answer: (164, 810, 1677)/94.

Ff. 193v-194r, prob. 41. 3 men find a purse and want to buy a horse, giving: x + y + p/3  =  h, y + z + p/5  =  h, z + x + p/4  =  h. If T = x + y + z, one gets T + 47p/60 = 3h and the solution space is actually two dimensional. He assumes p = 60, h = 47 and then this problem reduces to prob. 39 (cf in 7.R.1). Cf Lucca 1754 61r-61v; della Francesca 40v.

Calandri, Raccolta. c1495. Prob. 25, pp. 24 25. I (1/3, 1/5) with horse worth 60. Answer:  (300, 160)/7. Cf Calandri.

Riese. Rechnung. 1522. 1544 ed. -- pp. 91 92 & 94 95; 1574 ed. -- pp. 61v 62v & 63v 64r.

I (⅓, ¼) with horse worth 15. Answer: (120, 135)/11. Cf al-Karkhi I-42.

I (⅔, ¾) with horse worth 39. Answer: (52, 39)/2.

II (½, ⅓, ¼) with cow(?) worth 200. Answer: 64, 72, 84. Cf Lucca 1754 58r.

Tonstall. De Arte Supputandi. 1522. Pp. 246-248. I-(½, ⅓, ¼) with common value 20. Cf Gherardi 36-37.

Riese. Die Coss. 1524. Many examples, including the following.

No. 31, pp. 44 45. 3 men buy a horse worth 100, II (½, ⅓, ¼). Cf Lucca 1754 58r.

No. 47, pp. 46 47. Same with horse worth 17, I (½, ⅓, ¼). Cf Lucca 1754 58r.

No. 48, p. 47. 3 men buy 3 horses leading to: a + b + c/5  =  12,   a/2 + b + c  =  18,  a + b/3 + c  =  16.

No. 120, p. 56. Same as no. 31.

No. 121, p. 56. 3 men buy horse worth 100, II (3/4, 4/5, 5/6). Answer:  (510, 520, 475)/9.

No. 122, p. 56. Same, I (½, ⅓, ¼). Says his friend Hans Conrad learned this from a Dominican(?) monk named Aquinas. Cf Gherardi 36-37.