Pp. 23-34, Quest. VIII. 5x + 7y + 9z = 93256. (13807365, 13801148) solutions of which he gives the positive ones. Again, he does this two different ways. Pp. 34-35, Quest. IX. 3x + 5y + 9z = 1849. (12710, 12546) solutions. He outlines the method, which leads to adding three arithmetic progressions, but he makes a simple error which leads to a considerably smaller number. Pp. 36-37, Quest. X. 3x + 5y + 20z = 1849. (5766, 5612) solutions. He outlines the method and it does yield the number of positive solutions. Pp. 38-39, Quest. XI. 3x + 5y + 17z = 1849. (6794, 6613) solutions. He outlines the method and it does yield the number of positive solutions. Pp. 39-44, Quest. XII. 2x + 3y + 5z + 30w = 100003. (185312986853, 185090752407) solutions. He sketches the method and says it all adds up to 160190378249 positive solutions. I have not tried to locate his mistake, but one can estimate the number of solutions of ax + by + cz + dw = e as e3/6abcd which is 185201 x 106 here, so it is clear that his answer is wrong. Indeed, even his final addition is incorrect -- I get 160412356049. He finds that the number of solutions for w = 3333, 3332, ... is a quadratic and sums this by Newton's interpolation formula. I did some calculations and found that the number of positive solutions for w = 3333, 3332, ... is 1, 24, 77, 160, 273, ..., whose first differences are 23, 53, 83, ... and second differences are constant at 30. Hence the total number of positive solutions is obtained by adding 3333 terms, which is given by 0 + 1 * 3333 + 23 * 3333·3332/2 + 30 * 3333·3332·3331/6 and this gives the answer I found previously. For reasons which I haven't tried to determine, Dodson gets 0 + 1 * 3333 + 20 * 3333·3332/2 + 26 * 3333·3332·3331/6. Pp. 46-63, Quest. XIV. 3x + 57 + 19z + 143w = 91306. (3121604438, 3104216955) solutions. He outlines the method but I haven't tried to see if it does yield the number of positive solutions.
Euler. Algebra. 1770.
II.I, pp. 302 310.
Art. 8: Question 5. Men pay 19, women 13, to total 1000. He gives the general solution and all (4, 4) solutions. Art. 9: Question 6. Buy horses worth 31 and oxen worth 21 to cost 1770. General solution and all (3, 3) solutions. Art. 15: Question 9. Men pay 25, women pay 16. All together the women pay 1 more than the men. General solution and first few examples. Art. 16: Question 10. Horses cost 31 and oxen 20. All together, the oxen cost 7 more than the horses. General solution and first few examples. Art. 17 19. General rule for bp = aq + n. Applies to the above problems and Chinese Remainder problems.
II.II, pp. 311 317.
Art. 25: Question 1. 30 for 50 at 3, 2, 1. (Men, women, children.) Gives all (11, 9) answers. Art. 26: Question 2. 100 for 100 at 7/2, 4/3, ½. This has (4, 3) solutions. He gives (3, 3) answers and mentions 0, 60, 40 as another answer. Art. 27. Discusses when such problems are impossible, e.g. 100 for 51 at 7/2, 4/3, ½. Art. 28: Question 3. Combine silvers of quality 7, 11/2, 9/2 ounces per marc (= 8 ounces) to produce 30 marcs of quality 6. I.e. x + y + z = 30 and 7x + 11y/2 + 9z/2 = 6 x 30 or 30 for 180 at 7, 11/2, 9/2. He gives all (5, 3) solutions. Art. 29: Question 4. 100 for 100 at 10, 5, 2, ½. (Oxen, cows, calves, sheep.) Gives all (13, 10) answers. Art. 30: Question 4. 3x + 5y + 7z = 560 and 9x + 25y + 49z = 2920. Gives all (2, 2) answers.
II.III: Questions for practice, p. 321.
No. 4. Old guineas worth 21½ shillings and pistoles worth 17s to make 2000s. Gives all (3, 3) solutions. No. 5. 20 for 20 at 4, ½, ¼. = Pacioli 18. Gives (1, 1) of the (2, 1) solutions. No. 7. Can one pay £100 (= 2000s) with guineas (= 21s) and moidores (= 27s)? = Simpson 4. No. 8. How to pay 1s to a friend when I only have guineas (= 21s) and he only has louis d'or (= 17s)? I.e. 21x 17y = 1 with x, y positive. He gives the least solution. No. 9. Same as Simpson XIV.
Mr. Moss, proposer and solver. Ladies' Diary, 1773-74 = T. Leybourn, II: 374-376, quest. 658. Pay 50£ (= 1000s) with pistoles (17s), guineas (21s), moidores (27s) and six-and-thirties (36s). (529, 412) solutions, of which the positive ones are given.
Ozanam-Montucla. 1778. [This is one of the few topics where the 1725, 1778 and 1803 editions vary widely -- see each of them.]
Prob. 9, 1778: 195. Pay 2000 with 450 coins worth 3 and 5, i.e. 450 for 2000 at 3, 5. Prob. 13, 1778: 204-205. This is a complicated problem on the number of ways to make change -- see third section below. He breaks it up into silver and copper coins as given in the first two sections. Pay 60 with coins worth 60, 24, 12, 6. There are (13, 0) solutions -- he says 13. Pay 6k with coins worth 2, 1½, 1, ½, ¼, for k = 1, 2, ..., 10. There are (155, 2), (1292, 194), (5104, 1477), (14147, 5615), (31841, 15236), (62470, 33832), (111182, 65759), (183989, 116237), (287767, 191350), (430256, 298046) solutions. He gives the number of non-negative solutions in each case, but he gives four wrong values: 62400, 183999, 287777, 430264. Pay 60 with coins worth 60, 24, 12, 6, 2, 1½, 1, ½, ¼. There are (1814151, 0) solutions -- he says 1813899, which corresponds to using all but the third of the above wrong values to do the final calculation. Presumably the third wrong value is a misprint, rather than an error. Prob. 23, 1778: 214-216. A must pay B 31, but A has only pieces worth 5 and B has only pieces worth 6, i.e. 5x - 6y = 31. Gives the general solution.
The following are contained in the Supplement and have no solutions given.
Prob. 45, 1778: 433. 120 for 2400 at 12, 24, 60. (Paying with coins.) (21, 19) solutions. Prob. 51, 1778: 434. A, B and C have 100, but nine times A plus 15 times B plus 20 times C equals 1500, i.e. 100 for 1500 at 9, 15, 20. Notes that this problem, and other similar problems, have several solutions and one should find them all. This has (10, 9) solutions.
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