Prob. 52, 1778: 434-435. 120 for 20 at 3, 2, ½. (Hares, pheasants, quails.) [There must be a misprint here as this is clearly impossible. Perhaps it should be 20 for 20, which would = Alcuin 32. ??]
Prob. 52, 1778: 434-435. 120 for 20 at 3, 2, ½. (Hares, pheasants, quails.) [There must be a misprint here as this is clearly impossible. Perhaps it should be 20 for 20, which would = Alcuin 32. ??]
Prob. 57, 1778: 435. Pay 24 livres with demi-louis, pieces worth 6 livres and pieces worth 3 livres. [I think a demi-louis is 10 livres, so this would be 10a + 6b + 3c = 24.] This has (5, 0) solutions.
Prob. 58, 1778: 435-436. 10a + 6b + 3c + 2d + e + ½f = 24. [This is like the preceding and I have assumed that the demi-louis is 10 livres.] This has (1178, 2) solutions.
Bonnycastle. Algebra. 1782. Pp. 135-137 give a number of problems of finding some or all the integral solutions of one linear equation in two or three variables.
P. 135, no. 3. Same as Simpson 2. All answers given.
P. 136, no. 8. Same as Simpson 10. He says there are 60 solutions, but does not give them.
P. 136, no. 9. Same as Simpson 4.
P. 136, no. 10. Same as Simpson 3.
P. 136, no. 11. Same as Bachet's first.
P. 136, no. 12 (1815: p. 158, no. 10). Same as Euler II.III.8. He gives the least solution.
P. 136, no. 14. Pay £351 (= 7020s) with guineas (21s) and moidores (27s). Asks for the fewest numbers of pieces, and implies that there are 36 answers, but there are (38, 37) answers. See p. 206, no. 49.
P. 137, no. 15. Mix wines worth 18, 22, 24 per gallon to make 30 gallons worth 20, i.e. 30 for 600 at (18, 22, 24). (6, 4) answers -- he gives the 4 positive ones.
P. 137, no. 16. Pay £100 (=2000s) in crowns (5s), guineas (21s) and moidores (27s). He says there are 70734 answers. This is intended to be Simpson XIV, but that had £1000 (!). This has only (725, 691) solutions.
P. 206, no. 49 (in 1805, no. 48 in 2nd ed., 1788.)
"With guineas and moidores, the fewest, which way,
Do the pieces amount to? -- my fortune's to come."
This is the same problem as p. 136, no. 14, but here he says the answer is 9 guineas and 233 moidores (which was 257 in 2nd ed., 1788, but should be 253), so he is ignoring the case with 0 guineas and 260 moidores -- or this is a misprint. He says there are 37 solutions -- there are (38, 37) solutions.
Hutton. A Course of Mathematics. 1798? Prob. 19, 1833: 221; 1857: 225. Pay £120 (= 2400s) with 100 coins using guineas (= 21s) and moidores (= 27s), i.e. 100 for 2400 at 27, 21.
John Stewart. School exercise book of 1801 1802. Described by: W. More; Early nineteenth century mathematics; MG 46 (No. 355) (Feb 1962) 27 29. "Having nothing on me but guineas and having nothing on him but pistoles, I wish to pay him a shilling." = Euler II.III.8. Least solution given.
Ozanam-Hutton. 1803. [This is one of the few topics where the 1725, 1778 and 1803 editions vary widely -- see each of them.]
Prob. 9, 1803: 192; 1814: 167; 1840: 86. Pay £2000 with 4700 half-guineas and crowns, i.e. 4700 for 40000 at 10½, 5. No solution.
Prob. 23, 1803: 209-210. Prob. 22, 1814: 181-183; 1840: 94. A must pay B 31, but A has only pieces worth 7 and B has only pieces worth 5, i.e. 7x - 5y = 31. Gives the general solution.
The following are contained in the Supplement and have no solutions given.
Prob. 45, 1803: 426; 1814: 361; 1840: 186. Pay £100 (= 2000s) with guineas (= 21s) and pistoles (= 17s), i.e. 21a + 17b = 2000. = Simpson 3.
Prob. 51, 1803: 427; 1814: 362; 1840: 187. Same as Ozanam-Montucla prob. 51.
Prob. 52, 1803: 427; 1814: 362; 1840: 187. 100 for 100 at 7/2, 4/3, ½. (Calves, sheep, pigs.) This has (4, 3) solutions. = Euler II.II.26.
Prob. 57, 1803: 428; 1814: 362; 1840: 187. Divide 24 into three parts a, b, c so that 36a + 24b + 8c = 516, i.e. 24 for 516 at 36, 24, 8. This has (3, 3) solutions.
Bonnycastle. Algebra. 10th ed., 1815.
P. 158, no. 9. Pay £20 (= 400s) with half-guineas (= 10½ s) and half-crowns (= 2½ s). This gives 21x + 5y = 800. (8, 7) solutions -- he says there are 7.
P. 158, no. 11. Mix spirits worth 12, 15, 18 per gallon to make 1000 gallons worth 17. This has no integral solutions -- he gives one fractional solution.
P. 228, no. 21. Pay £100 (= 2000s) with 7s pieces and dollars (worth 4½ s). I.e. 14x + 9y = 4000. I find (31, 31) solutions, he says there are 21 -- a misprint?
P. 230, no. 34. Spend 28s (= 336d) on geese worth 52d and ducks worth 30d. (1, 1) solution, which he gives.
Augustus De Morgan. Arithmetic and Algebra. (1831 ?). Reprinted as the second, separately paged, part of: Library of Useful Knowledge -- Mathematics I, Baldwin & Craddock, London, 1836. Arts. 310-317, pp. 103-110, cover indeterminate problems in general.
Art. 312, pp. 105-106. Pay 2£ 16s with crowns (= 5s) and three shilling pieces. Describes the (4, 4) solutions. Observes that negative solutions correspond to refunding. Notes that paying in crowns (= 5s) and half-sovereigns (= 10s) is impossible.
Art. 314, pp. 107-108. Discusses 4x + 9y + 10y = 103 with no context. Describes the (17, 13) solutions.
Art. 315, p. 108. Discusses 14x + 11y + 9z = 360; x + y + z = 30 without any context and describes how to solve such problems in general. Gives (4, 3) of the (5, 3) solutions, omitting 18, 0, 12.
Bourdon. Algèbre. 7th ed., 1834.
Art. 127, pp. 221-222. Pay 78 fr. with pieces worth 5 fr. and 3 fr. Sees that the number of 5 Fr. pieces must be divisible by 3 and finds all (6, 5) solutions.
Art. 138, question 3, pp. 238. = Euler II.II.28.
Art. 141, question 7, pp. 243-244. = Euler II.II.29.
Unger. Arithmetische Unterhaltungen. 1838. Pp. 145-172 & 259-262, nos. 541-645. He treats the topic at great and exhausting length, starting with solving ax = by, then x + y = c, then ax + y = c, then ax = by + 1 and ax = by + c, also phrased as ax c (mod b). He then does Chinese Remainder problems, but with just two moduli. He continues with ax + by = c, initially giving just one solution, but then asking how many solutions there are. He varies this in an uncommon way -- solve x + y = e with x a (mod b) and y c (mod d). He also varies the problem by letting a, b be rationals. After 20 pages and 90 problems, he finally gets to one problem, no. 631, of the present type. He then goes on to ax + by = c (x + y), but returns with 12 problems of our type, no. 634-645. Of these, 640 and 645 have an answer with a zero, but he only gives the positive answers. None of these problems are the same as any others that I have seen.
