Pp. 224 225 (S: 332-333). Four men & purses: p, p+10, p+13, p+19, with A = (2, 3, 4, 5). Pp. 225 226 (S: 333-335). Four men, one purse, giving w + x + p = 2y, etc. P. 227 (S: 335-336). Three men, one purse, giving x + y + p = 2z, etc. P. 227 (S: 336). Four men, one purse, giving w + x + p = 2y + z, etc. Pp. 227 228 (S: 336-337). Five men, one purse, giving v + w + p = 2 (x + y + z), etc., with constants 2, 3, 4, 5, 6. Answer: 22, -9, 57, 12, 71; 267 (the text has 7 instead of 71). "... quare hec questio est insolubilis, nisi ponamus, secundum hominem habere debitum 9, ..." [... therefore this problem is unsolvable unless we put the second man to have a debit of 9, ...]. See Sesiano. P. 284 (S: 405). III (31, 34, 37, 27). P. 285 (S: 405-406). III (31, 34, 37, 39, 27). On pp. 284-285 & 286, he also considers problems like IV-(a, b, c, d) (cf. Bakhshali) and notes that some are inconsistent (e.g. for a, b, c, d = 27, 31, 34, 37) and others are indeterminate. Pp. 302 303 (S: 426-427). Four men, one purse, giving equations w + x + p = 3/2 y, etc., with constants 3/2, 9/4, 16/5, 25/6. Answer: 8665, 5682, 12718, 10280; 4730. Pp. 326 327 (S: 456-458). Problem of pp. 218 220 done by false position. Pp. 330-331 (S: 461-462). III-(75, 70, 67, 64, 54, 50). P. 333 (S: 465). Problem of pp. 214 216 done by false position. Pp. 346 347 (S: 480-481). Three men & purses of values 18, 16, 20, giving x' + 18 = 3y', y' + 16 = 4z', z' + 20 = 5x'. By setting x' = x 11, y' = y 7, x' = z 9, he converts to the ass & mule problem (7.R) on pp. 344 346. Pp. 349 352 (S: 484-487). Four men, one purse, giving w + p = 2 (x + y), etc., with constants 2, 3, 4, 5. Answer: 1, 4, 1, 4; 11. "... hec questio insolubilis est, nisi concedatur, primum hominem habere debitum, ..." [... this problem is not solvable unless it is conceded that the first man can have a debit, ...]. See Sesiano.
Fibonacci. Flos. c1225. In Picutti, pp. 316-319, numbers IV-V.
Pp. 238-239: De quatuor hominibus et bursa ab eis reperta, questio notabilis. Described as the second of the problems that Fibonacci sent to Frederick II. Same as Fibonacci, pp. 349 352, though he doesn't cite this. "... hanc quidem questionem insolubilem esse monstrabo, nisi concedatur, primum hominem habere debitum: ...." See Sesiano. Pp. 239-240: De eadem re. Does the same problem with multipliers 4, 5, 6, 7. Answer: -1, 6, 1, 6; 29. Implies a general solution for multipliers k 2, k 1, k, k + 1 is -1, k, 1, k; k2 - k - 1.
Jordanus de Nemore. De Numeris Datis. c1225. Critical edition and translation by Barnabas Hughes. Univ. of Calif. Press, Berkeley, 1981. Prob. II 24, pp. 150 151. General version of type I with purse given. Example: I (1/9, 1/3, 3/5, 1) with purse 6. Answer: 2, 14, 24, 34.
Ibn Badr = Abenbéder = Abu ‘Abdallah Muhammad (the h should have an underdot) ibn ‘Umar ibn Muhammad (the h should have an underdot). c1225. Arabic text with Spanish translation by José A. Sánchez Pérez as: Compendio de Álgebra de Abenbéder; Centro de Estudios Históricos, Madrid, 1916. Tercer problema análogo, pp. 109-111 (pp. 70-71 of the Arabic). (4, 7). Answer: (8, 5; 27)/9.
BR. c1305. No. 61, pp. 84 87. (7, 11). Answer given is 12, 8; 76, but should be 8, 12; 76.
Gherardi. Libro di ragioni. 1328. P. 53. 3 men find a purse. I (2, 3, 4). Answer: 7, 17, 23; 73.
Lucca 1754. c1330. F61r, pp.139 140. II (2, 3, 4, 5). Answer: 33, 76, 65, 46; 119. (= Fibonacci, pp. 218 220.)
Bartoli. Memoriale. c1420. Prob. 4, f. 75r (= Sesiano, pp. 136-137 & 147. I (2, 3, 4). Answer: 7, 17, 23; 73. (= Iamblichus's 1st.)
Pseudo-dell'Abbaco. c1440. Prob. 125, p. 100. I (2, 3, 4). Answer: 7, 17, 23; 73. (= Iamblichus' 1st problem.)
AR. c1450. Prob. 113, 159, 227. Pp. 63 64, 76, 168 169, 217.
113: I (1, 2, 3), answer: 2, 10, 14; 22. 159: I (4, 10), answer: 5, 11; 39. 227: I (1, 2, 3), answer: 1, 5, 7; 11. Cf. 113.
Correspondence of Johannes Regiomontanus, 1463?-1465. Op. cit. in 7.P.1.
P. 238, letter from Bianchini, 5 Feb 1464. Query 8: III (42, 54, 30). Pp. 259 & 291, letter to Bianchini, nd [presumably 1464]. P. 259 gives the answer: 21, 9, 33. P. 291 is part of Regiomontanus's working for this letter where he solves the problem by assuming the first has x, so the second has 30 - x and the third has 54 - x, hence 84 - 2x = 42.
Benedetto da Firenze. c1465.
P. 67: III (12, 14, 10). P. 68: III (10, 8, 15, 12).
Chap. 22: "... huomini che ànno danari et trovano borse di danari", pp. 181 183.
I (2, 3), answer: 3, 4; 5. (= Mahavira v. 244.) II (2, 3, 4), answer: 9, 16, 13; 23. (= Fibonacci, p. 216.) I (2, 3, 4), answer: 7, 17, 23; 73. (= Iamblichus's first.)
Muscarello. 1478. Ff. 65r-65v, pp. 172-173. Men find a purse, I (3, 4, 5). Answer: 7, 13, 17; 83.
della Francesca. Trattato. c1480.
F. 42v (105-106). This is a type I problem with four values and no purse, except the last value is specified. The multipliers are 7/12 (given as 1/3 + 1/4), 9/20, 11/30 and z = 8. So the first equation is w = (7/12)(x+y+z). We could interpret 7z/12 as a purse of negative value, but it is easier to write the first equation as w = (7/12)(T-w), so w = (7/19)T which leads to T = (7/19 + 9/29 + 11/41)T + 8. Answer: (66584, 56088, 48488, 9568)/1196. F. 120v (257). III-(20, 30, 40). Answer: 25, 25, 5.
Chuquet. 1484.
Triparty, part I. FHM 80-81 & 83-85. Sesiano cites pp. 642 & 646 of Chuquet, ??NYS.
III-(120, 90, 80, 75). Answer: (5, 95, 125, 140)/3. Not mentioned by Sesiano; English in FHM 80, where the next two cases are just given as formulae with solutions. III-(120, 90, 80, 75, 72). Answer: (-43, 77, 117, 137, 149)/4. Sesiano notes that this is the same as the solution of the problem in 7.R.2 in the Provençale Arithmétique, c1430, and in Pellos, 1492.
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