7. arithmetic & number theoretic recreations a. Fibonacci numbers

I (1/5 + 2/3, 1/480 + 1/6 + 2/3, 1/638 + 1/6 + 2/3, 1/420 + 1/7 + 2/3, 1/810 + 1/27 + 1/10 + 2/3). Answer:  3, 228, 231, 348, 378; 1030.

Pp. 250 251 (S: 364-365). Five men, giving v + w + ½ (x + y + z)  =  ...  =  h, with constants 1/2, 1/3, 1/4, 1/5, 1/6. Answer:  58, 19, 148, 49; 163.

Pp. 251 252 (S: 365-366): Questio insolubilis [An unsolvable problem]. Four men, giving equations w + x + ½ (y + z)  =  ...  =  h, with constants 1/2, 1/3, 1/4, 1/5. This is inconsistent unless h = 0. He then varies the constants to 1/2, 3/7, 3/11, 5/13, which gives answer:  5, 6, 7, 9; 19.

Pp. 252 253 (S: 366-367). Seven men, giving t + u + v + ½ (w + x + y + z)  = ...  =  h, with constants ½, ⅓, ..., ⅛. Answer:  507, 171,  9, 1347, 451, 131, 1431; 2349. "... quare tercius homo habet debitum ipsos 9, vel hec questio est insolubilis: sit itaque solubilis cum debito tercii hominis; ..." [... therefore the third man has a debit of 9 bezants, or this problem is unsolvable; ...]. He then varies the constants to 1/3, 1/4, ..., 1/9, with answer: 1077, 717, 489, 1637, 997, 657, 1749; 3963. See Sesiano.

Pp. 253 254 (S: 367-368). Two men and two horses of values h₁, h₂ = h, h+2. i-th says "If I had a_i of what the rest of you have, then I could buy the i-th horse", with constants ⅓, ¼. (This is the same as on pp. 235 236 above, but only because n = 2.)

P. 254 (S: 368). Same with 3 men, 3 horses worth h, h+2, h+4 and constants 1/3, 1/4, 1/5. I.e. I-(1/3, 1/4, 1/5) with h_i = h, h+2, h+4. Answer: 7, 13, 17; 17.

Pp. 254 257 (S: 369-371). Same with 4 men, 4 horses worth h, h+3, h+7, h+12 and constants 1/3, 1/4, 1/5, 1/6. I.e. I-(1/3, 1/4, 1/5, 1/6) with

h_i = h, h+3, h+7, h+12. Answer: ( 4, 13, 27, 41; 23)/2. "Unde hec questio cum hiis iiii or positis residuis solui non potest, nisi primus homo haberet debitum." [Whence this problem with these IIII posed residues can be solved with the first man having a debit. (Sigler), but it would be more literal to have Whence this problem with these IIII posed residues can not be solved unless the first man has a debit.]. He later gives another solution: (82, 193, 265, 325; 343)/6, by choosing a larger value for h. The margin has 32 1/2 for 32 1/6 which I have converted to 193/6. See Sesiano.

Pp. 257 258 (S: 371-372). Four men, one horse, giving w + x/2 + y/3 + z/4  =  x + y/3 + z/4 + w/5  =  ...  =  h. Answer:  105, 168, 210, 240; 319.

Pp. 327 329 (S: 458-460). II (1/2, 1/3, 1/4, 1/5, 1/6) done by false position. Answer:  456, 530, 573, 592, 645; 721 -- the text has 529 instead of 592. Sesiano notes that negatives are used in one of the false positions.

Pp. 334 335 (S: 466-467). Problem of pp. 245 248 done by false position.

Pp. 336-338 (S: 469-470). (⅓, ¼) with h_i = 14, 17. Answer:  (100, 162)/11.

P. 338 (S: 470-471). II-(1/3, 1/4, 1/5) with h_i = 14, 17, 19, done by false position. Answer:  (595, 777, 1040)/61.

Pp. 338-339 (S: 471-472). I-(1/3, 1/4, 1/5) with h_i = 14, 17, 19, done by false position. Answer:  (241, 594, 783)/50.

Pp. 347 349 (S: 481-483). Three men, one horse, giving x + y/2 + z/3  =  y + z/4 + x/5  =  z + x/6 + y/7  =  h. Answer:  1530, 3038, 3540; 4229. Done two ways.

P. 349 (S: 484). Four men, one horse, giving w + x/2 + y/3 + z/4  =  x + y/4 + z/5 + w/6  =  y + z/6 + w/7 + x/8  =  z + w/8 + x/9 + z/10  =  h. Answer:  8569848, 21741336, 26955060, 29657460; 35839901.

Fibonacci. Flos. c1225. In: Picutti, pp. 312-316 & 320-326, numbers III, VI & VII.

Pp. 236-238: De quinque numeris reperiendis ex proportionibus datis. Five values: v + (w + x + y)/3 = w + (x + y + z)/3 = .... Answer: 7, 10, 19, 25, 28; 34.

Pp. 240-242: No heading -- paragraph begins: Item de mode predicto extraxi .... I (1/3, 1/4, 1/5) with h_i = 14, 17, 19. Answer:  (241, 594, 783)/50.

Pp. 242-243: De quatuor hominibus bizantios habentibus. He refers to Liber Abaci, apparently to pp. 338-339. I (1/2, 1/3, 1/4, 1/5) with horses worth 33, 35, 36, 37. Answer:   3, 18, 25, 29. "... hanc insolubilem esse sub posita conditione." On p. 243, he states that if the values of the horses are  181, 183, 184, 185, then the answer is 1, 94, 125, 141. See Sesiano, who notes that for h_i = h, h+2, h+3, h+4, the smallest positive integral solution is that given by Fibonacci.

Fibonacci. Epistola. c1225. In Picutti, pp. 338-340, numbers XV & XVI.

Pp. 250-251: Modus alius solvendi similes questiones. I (1/2, 1/3, 1/4, 1/5, 1/6) with horses worth 12, 15, 18, 20, 23. Answer:  (4938, 7428, 10161, 11268, 15760)/721.

Pp. 251-252: Investigatio unde procedat inventio suprascripsit. II (1/2, 1/3, 1/4, 1/5, 1/6) with horses worth 12, 15, 18, 20, 23. Answer:  ( 5316, 1479, 4532, 6157, 7920)/394. He says "tunc questio esset insolubilis, nisi concederetur, primum habere debitum; quod debitum esset [5316/394]." See Sesiano.

Jordanus de Nemore. c1225. Op. cit. in 7.R.1.

Prob. II 25, p. 151. General version of type II with value of horse given. Example: II (1/2, 1/3, 1/4, 1/5) with horse worth 119. Answer: 75, 88, 93, 104.

Prob. II 26/28, pp. 152 155. General version of type I with value of horse given. Example in II 26: I (3, 13/4, 25/7, 4) with horse worth 28. Answer: 1, 2, 3, 4. Example in II 27: I (1/2, 1/3, 1/4, 1/5) with horse worth 37. = Fibonacci

245 248. Answer:  1, 19, 25, 25.

II 28 is II 26 done in a different way.

BR. c1305.

No. 6, pp. 26 27. Like Fibonacci's pp. 242 243 with constants 1/5, 1/7, 1/9 and h = 100. Method is wrong, but the answer is right:  (2700, 2800, 3000)/61.

No. 7, pp. 26 29. Buying a ship, I-(⅓, ¼) with ship worth 100. Answer:  (800, 900)/11. Cf al-Karkhi I-24.

No. 8, pp. 28 29. Buying a business, I-(2/3, 3/5) with business worth 100. Answer:  (215, 258)/9.

No. 50, pp. 66 69. Same as no. 6 with constants (1/3, 1/4, 1/5) and h = 11. Answer:  45, 48, 54.