Prob. 24, question 8, p. 180. 41 for 40 at 4, 3, ⅓. (Men, women, children.) = Bachet's first. (1, 1) solutions which he gives. Prob. 50, pp. 255 256. 100 for 100 at 9, 1, ½, 3. "Ce problême est capable d'un grand nombre de résolutions; ...." It has (73, 46) solutions -- he gives (3, 3).
Dilworth. Schoolmaster's Assistant. 1743. Part IV: Questions: A short Collection of pleasant and diverting Questions, p. 168. Problem 3. 20 for 20 at 4, ½, ¼. (Pigeons, larks, sparrows.) This has (2, 1) solutions, none given. = Pacioli 18.
Simpson. Algebra. 1745. Section XIII, quest. 2-8, 10-12, pp. 170-181 (1790: prob. II VII, IX, XI XV, pp. 183 200).
2. 9x + 13y = 2000. (17, 17) answers which he gives. 3. Pay £100 (= 2000s) in guineas (= 21s) and pistoles (= 17s), i.e. 21x + 17y = 2000. (6, 6) solutions which he gives. 4. Pay £100 (= 2000s) in guineas (= 21s) and moidores (= 27s), i.e. 21x + 27y = 2000. But 2000 is not a multiple of three, so there are no solutions, as he notes. 5. Buy sheep at 17 and oxen at 140 to cost 2000. (1, 1) solution which he gives. 6. Men pay 42, women 16, to make 396. (1, 1) solution which he gives. 7 (1745 only). 20 for 16 at 2, 1, ¼. (Loaves of bread of different values.) Gives (1, 1) of the (2, 2) solutions. He gets the equation y = 14 - 2x + (2-x)/3 and neglects to allow x = 2 in it. 8 (1745 only). 20 for 20 at 3, 2, ½. (Expenditure of men, women, children.) Gives (1, 1) of the (2, 1) solutions. = Alcuin 32. 10 (1790: XI). 5x + 7y + 11z = 224. This has (72, 59) solutions. He gives (60, 60), but he has two erroneous solutions for z = 14 instead of one. 11 (1790: XII). 17x + 19y + 21z = 400. (13, 10) solutions. In 1745, there is an algebraic mistake and he gets (9, 9) solutions. This is corrected to (10, 10) in 1790. 12 (1745 only). Pay £20 (= 400s) in pistoles (= 17s), guineas (= 21s) and moidores (= 27s). Gives (9, 7) of the (9, 7) solutions, which is the only time here that he gives solutions with zeroes and is unusual for the time. Cf. prob. XIV in the 1790 ed. The following are in the 1790 ed. VII 12 for 12 at 2, 1, ¼. (Loaves of bread of different values.) (2, 1) solutions, he gives (1, 1). IX. 87x + 256y = 15410 -- find the least solution. In fact, there is only (1, 1) solution. XIII. 7x + 9y + 23z = 9999. This has (34634, 34365) solutions -- he describes the 34365 positive solutions. XIV. Pay £1000 (= 20000s) in crowns (= 5s), guineas (= 21s) and moidores (= 27s), i.e. 5x + 21y + 27z = 20000. This has (70734, 70395) solutions. He describes all 70734. Cf. Euler II.III.9; Bonnycastle, 1782, no. 16. XV. 12x + 15y + 20z = 100001. There are (1388611, 1388611) solutions which he describes.
Dodson. Math. Repository. (1747?); 1775. He also has several mixture problems and some simple problems which are mentioned at the entry for vol. II, below.
P. 16, Quest. XLI. Pay £50 (= 1000s) with 101 coins worth 21/2 s and 5s. P. 139, Quest CCXXIII. 20 for 20 at 4, ½, ¼ (geese, quails, larks). There are (2, 1) answers. He gives the positive one. = Pacioli 18. P. 140, Quest CCXXIV. Pay £100 (= 2000s) with guineas (= 21s) and pistoles (= 17s). Gives all (6, 6) solutions. = Simpson 3. P. 154, Quest. CCXXXVIII. 41 persons spent 40s; men spent 4s, women 3s, children ⅓ s. I.e. 41 for 40 at 4, 3, ⅓. He gives the (1,1) answer. = Bachet's first. Pp. 331-336, Quest CCI. Divide 200 into five whole numbers a, b, c, d, e, so that 12a + 3b + c + e/2 + z/3 = 200. I.e. 200 for 200 at 12, 3, 1, ½, ⅓. He finds 6627 answers. There are (8331, 6627) answers. = Tartaglia 45.
Les Amusemens. 1749. Prob. 164, p. 310. 20 for 60 at 6, 4, 1. (Men, women, valets.) (3, 2) solutions -- he gives (1, 1).
James Dodson. The Mathematical Repository. Vol. II. Containing Algebraical Solutions of A great Number of Problems, In several Branches of the Mathematics. I. Indetermined Questions, solved generally, by an elegant Method communicated by Mr. De Moivre. II. Many curious Questions relating to Chances and Lotteries. III. A great Number of Questions concerning annuities for lives, and their Reversions; wherein that Doctrine is illustrated in a Multitude of interesting Cases, with numerical Examples, and Rules in Words at length, for those who are unacquainted with the Elements of these Sciences, &c. J. Nourse, London, 1753. On pp. 1 - 63, he deals with 13 examples of determining the number of solutions of a linear equation in two (5 cases), three (6 cases) or four (2 cases) unknowns. Because of the unusual extent of this, I will describe them all. (I had to completely revise my programs for counting the number of solutions in order to deal with these. The revision introduced double precision values for the counts and a way of computing the number of solutions for the two variable problem which speeded up the programs up by a factor of about 100, but one problem still took 2½ days!) P. 1 has a subheading: The Solution of indetermined Questions in Affirmative Integers, communicated by Mr. Abraham De Moivre, Fellow of the Royal Societies of London and Berlin.
Pp. 1-3, Quest. I. 35x + 43y = 4000. (3, 3) solutions which he gives. P. 4, Quest II. 71x + 17y = 1005. = p. 143, Quest CCXXVII of vol. I, 1775 version. (1, 1) solution which he gives. Pp. 5-6, Quest III. 21x + 17y = 2000 -- pay £100 with "guineas at 21, and pistoles at 17 shillings each" = Simpson 3 = P. 140, Quest. CCXXXVIII of vol. I, 1775 version. (6, 6) solutions, which he gives. Pp. 6-9, Quest IV. 5x + 8y = 1989. (50, 50) solutions which he gives. P. 9, Quest V. 3x + 5y = 173. (12, 12) solutions which he gives. Pp. 10-14, Quest. VI. 3x + 5y + 8z = 10003. (417584, 416250) solutions of which he gives the positive ones. Pp. 14-22, Quest. VII. 3x + 5y + 19z = 13051. (299440, 298204) solutions of which he gives the positive ones. He describes De Moivre's analysis and then does a different approach.
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