7. arithmetic & number theoretic recreations a. Fibonacci numbers

I 47: Usual version with two people. A = (⅓, ¼), B = (1, 2), D = (30, 70). Answer:  (156, 344)/5.

III 6: Usual version: A = (1/2, 1/3, 1/6), B = (1, 1, 1), C = (3, 2, 1). Answer:  33, 13, 1.

Fibonacci. Flos. c1225. Pp. 234-236: De tribus hominibus pecuniam comunem habentibus. = Fibonacci, below, pp. 293-294. = al Karkhi's III 6. Problem II in Picutti, pp. 310 312.

Cantor; op. cit. under Fibonacci in 7.P.1; 1863; p. 345, discusses the contest between Fibonacci and John of Palermo before Frederick II at Pisa in 1225 (or 1226) and says this problem is the third and last of the contest problems and this is how I read the text. However Picutti has this as the second problem. Licks, op. cit. in 5.A, says it was problem 5 in the contest. Vogel doesn't mention which problem it was.

Leonardo says he later found three further methods of solution, which are "in libro uestro, quem de numero composui, patenter inserui". Leonardo is here addressing the Emperor, so Vogel interprets 'libro uestro' as a book dedicated to the Emperor. Vogel interprets this as referring to the material in Liber Abbaci, so I have now dated the next entry as 1228 rather than 1202, although there is no mention of the contest or the Emperor in Liber Abbaci.

The solution here is different than below and Vogel calls this Fibonacci's third method, the shortest and cleverest, and which Fibonacci described as "exceedingly beautiful"'. Vogel notes the remarkable hybrid notations: XXX3 for 33; XXIII 1/1 for 23½; X 1/1 for 10½ in this 15C MS.

Fibonacci. 1228 -- see above entry. Pp. 293 297 (S: 415-420). Several versions. He often notes that the values x_i can be multiplied through by any value.

Pp. 293 294 (S: 415-417). A = (1/2, 1/3, 1/6), B = (1, 1, 1), C = (3, 2, 1). Answer:  33, 13, 1. (= Al Karkhi III 6.) Vogel calls this Fibonacci's first method and notes a minor variation, which Fibonacci may have intended as another method. Vogel notes some typographical errors.

P. 294 (S: 417). Same A, with B = (3, 2, 1) = C. Answer: 30, 15, 6.

P. 294-295 (S: 417-418. A = (1/3, 1/4, 1/5), B = (1, 1, 1), C = (3, 2, 1). Answer:  177, 92, 25.

P. 295 (S: 418). Same A, with B = (3, 2, 1) = C. Answer: 162, 96, 45.

Pp. 295 296 (S: 418-419). Same A, with B = (5, 4, 1), C = (3, 2, 1). Answer:  543, 296, 175. In this problem he computes 360 - 360 as 0 and 0/2  as 0.

Pp. 296 297 (S: 419-420). A = (1/2, 1/3, 1/6), B = (1, 1, 1), C = (5, 4, 1). Answer:  326, 174,  30. "Quare hec questio non potest solui, nisi solvatur cum aliqua propria pecunia tercii hominis ..." [Therefore this problem can only be solved with some smaller amount of money for the third man ...] He treats  30 as a debt of the fund to the third man who steals nothing, thereby losing 30. See Sesiano.

P. 297 (S: 420). Four men. A = (1/2, 1/3, 1/4, 1/5), B = (1, 1, 1, 1), C = (10, 9, 6, 5). Answer:  1034, 666, 300, 190.

Pp. 335 336 (S: 468-469). Problem on pp. 293 294 done by false position. Vogel calls this Fibonacci's second method, but there is a typographical error citing p. 235f.

Gherardi. Libro di ragioni. 1328. Pp. 54 56. A = (½, ⅓, ¼), B = (1,1,1), C = (6, 4, 3). Answer is 240 : 93 : 44. He gives 4/29 of these values by starting with x₁ + x₂ + x₃ = 52.

Lucca 1754. c1330. Ff. 60v 61r, pp. 138 139. (= al Karkhi III 6.) Gives some explanation, but Vogel says only the beginning makes sense.

Columbia Algorism. c1350. Prob. 5, pp. 34 35. (= al Karkhi III 6.) Gives only the answer with no explanation. Vogel's Introduction, p. 22, sketches the history.

Pacioli. Summa. 1494. Gives a number of versions.

F. 105r, prob. 11. A = (½, ⅓), B = (1, 1), D = (100, 100). Solution is (600, 800)/7.

F. 157v, prob. 77. = al-Karkhi III-6 with total 12. He then sketches the solution with total 60, but immediately has wrong values. He also seems to have changed some of the parameters -- his answers don't add up to 60, and he gives final values in the ratio 6 : 4 : 3.

Ff. 157v-158r, prob. 78. A = (1/4, 1/6, 1/5), B = (9, 7, 4), C = (6, 4, 3), with total 12. His working is correct until he has 2 158/1651 instead of 2 958/1651 for his unknown and then the erroneous value is used in the final steps. Answer should be (9639, 5526, 4650)/1651.

F. 158v, prob. 82' (the second problem numbered 82). Seems to be a discussion of modification of prob. 78 to have B = (1, 1, 1), C = (1/2, 3/10, 1/7), but he never uses the values in C and winds up giving the answers of prob. 77 for total of 564, namely 396, 156, 12.

F. 158v, prob. 83. A = (⅓, ¼), B = (1, 1), D = (15, 15).

Calandri, Raccolta. c1495. Same as Fibonacci, pp. 296 297. Calandri simply says it is "insolubile".

Tonstall. De Arte Supputandi. 1522. Pp. 244-245. Same as Pacioli, prob. 11.

Cardan. Practica Arithmetice. 1539. Chap. 66.

Section 90, ff. GG.vii.v - GG.viii.v (pp. 164-165). Same as Fibonacci, p. 295, first problem. Answer: 354, 184, 50.

Section 91, ff. GG.viii.v - HH.i.v (pp. 165-166). A somewhat similar situation, where the first two take money leaving the third with 5. Friend says the first is to give 10 and ⅓ of what he has left to the second and the second is then to give 7 and ¼ of what he has left to the third to make C = (3, 2, 1). This is determinate. Answer is x = 172, y = 39 and the total sum is 216.

Recorde. Second Part. 1552. Pp. 330-335: A question of partners, the ninth example. A = (¾, ⅓) or (⅓, ¾), the person with the larger x_i to give back ¾. B = (1, 1), D = (180, 120). Solution:  (1680, 1620)/11. The second person took the more money.

Buteo. Logistica. 1559.

Prob. 7, pp. 335-336. A = (½, ¼), B = (1, 1), C = (1, 1). He assumes total is 500, then answer is 300, 200.

Prob. 8, pp. 336-337. A = (½, ⅓, ¼), B = (1, 1, 1), D = (116, 116, 116). Answer:  144, 108, 96.

Vogel says that Clavius; Epitome Arithmeticae; Rome, 1595, pp. 249-252, gives a simple example with two persons and that then the problem vanishes from the literature.
7.Q. BLIND ABBESS AND HER NUNS -- REARRANGEMENT ALONG SIDES

OF A 3 x 3 SQUARE TO CONSERVE SIDE TOTALS
This is a kind of magic figure, except that here we generally have repeated values.

There are three trick versions of 6.AO which might be classified here or in 7.Q.2.

(12, 4, 5) -- Trick version of a hollow 3 x 3 square with doubled corners, as in 7.Q: Family Friend (1858), Secret Out, Illustrated Boy's Own Treasury.

Van Etten and Mittenzwey are the only inverse examples, where the total number remains fixed but the number on each side changes.

Shihâbaddîn Abû’l ‘Abbâs Ahmad [the h should have an underdot] ibn Yahya [the h should have an underdot] ibn Abî Hajala [the H should have an underdot] at Tilimsâni alH anbalî [the H should have an underdot]. Kitâb ’anmûdhaj al qitâl fi la‘b ash shaţranj [NOTE: ţ denotes a t with an underdot] (Book of the examples of warfare in the game of chess). Copied by Muhammed ibn ‘Ali ibn Muhammed al Arzagî in 1446. This is the second of Dr. Lee's MSS, described in 5.F.1, denoted Man. by Murray. Described by Bland and Forbes, loc. cit. in 5.F.1 under Persian MS 211, and by Murray 207 219.

Murray 280 says No. 46 49 give the problems of arranging 32, 36, 40, 44 men along the walls and corners so the total along each edge is 12.

Pacioli. De Viribus. c1500.