7. arithmetic & number theoretic recreations a. Fibonacci numbers

No. 25. (1, 1; 1, 2).

No. 26. (1, 1; 1, 3).

No. 27. (1, 10; 1, 20).

No. 28. (1, 2; 1, 5).

No. 29. II (1, 1; 1, 2; 1, 3).

No. 30. II (1, 1; 2, 2; 3, 3).

No. 63. (1, 1; 4, 2).

No. 64. (1, ½; 5, 3).

Cardan. Practica Arithmetice. 1539. Chap. 61.

Section 3, f. S.viii.v (p. 110). (5, 4; 4, 4).

Section 4, f. S.viii.v (p. 110). Variation leading to: x + ½y = 3 (y - ½y), y + ½x  =  7 (x - ½x), which he rightly states is impossible (unless x = y = 0).

Section 7, f. T.i.v (p. 111). Variation giving equations: x + ½y + 2 = 9 (y - ½y - 2), y + ⅓x + 3 = 3 (x - ⅓x - 3).

Recorde. Second Part. 1552. Pp. 322-324: The fourth example. (2, 4; 3, 1).

Tartaglia. General Trattato. 1556.

Book 16, art. 12 15, pp. 240v 241v. (6, 1; 9, 2). (7, 2; 13, 3). (30, 1; 30, 2). I (16, 1;  24, 2; 33, 3).

Book 17, art. 23, 34, 35, pp. 271v-272r & 273v-274v. II (32, 2; 38, 3; 50, 4; 76, 7). (24, 2; 42, 3). II-(34, 2; 52, 3; 80, 5).

Buteo. Logistica. 1559. Prob. 59, p. 264. i-th says "I have a_i times as much as the rest of you." with (a_i) = (1, 1/2, 1/5). This could be considered as I-(0, 1; 0, 1/2; 0, 1/5). This is indeterminate with general solution proportional to (3, 2, 1). He assumes x = 24 and gets y = 16, z = 8.

van Etten. 1624. Prob. 83 (76), parts a & c, pp. 90 92 (134 136). Ass & mule -- (1, 2; 1, 1) = Euclid. (10, 3; 10, 5), (2, 2; 2, 4) = Metrodorus.

Hunt. 1651. Pp. 280-281: Of the mule and the ass. (1, 2; 1, 1).

Schott. 1674. Ænigma V, pp. 553. (1, 1; 1, 2) = Euclid. Cites Euclid, Clavius and Lantz.

Wingate/Kersey. 1678?. Quest. 39, pp. 502-503. Ass and mule in Latin verse - cf Euclid. (1, 2; 1, 1)

Edward Cocker. Arithmetic. Ed. by John Hawkins. T. Passinger & T. Lacy, London, 1678. [De Morgan states "I am perfectly satisfied that Cocker's Arithmetic is a forgery of Hawkins" and then spends several pages detailing this charge and showing that the book is a rather poor compilation from several better books. However Ruth Wallis [Ruth Wallis; Edward Cocker (1632?-1676) and his Arithmetick: De Morgan demolished; Annals of Science 54 (1997) 507-522] has argued that De Morgan is wrong. Inspection of a 1st ed. at the Graves collection and a 3rd ed., 1680, at Keele shows no noticeable difference in the texts other than resetting which makes the book smaller with time -- all the editions seen have the same 32 chapters. The 1st and 3rd eds. seem to have identical pagination so I will not cite the 1680.] 1st ed., 1678 & 3rd ed., 1680, both T. Passinger & T. Lacy, London. = 33rd ed., Eben. Tracy, London, 1715. = Revised by John Mair; James & Matthew Robertson, Glasgow, 1787. Chap. 32, quest. 4. 1678: p. 333; 1715: p. 215; 1787: p. 186. (1, 5; 1, 1).

Wells. 1698. No. 104, p. 206. Ass & mule: (1, 1; 1, 2); (a, 1; a, 2).

Ozanam. 1725. De l'asne et du mulet, prob. 24, question 3, 1725: 176 178. Prob. 6, 1778: 189-190; 1803: 186-188; 1814: 162-163; 1840: 84-85. (1, 2; 1, 1). 1725 gives two versions and solutions in Latin verse. 1778 et seq. gives just one version and solution, but with slight differences, and refers to the Metrodorus problems in Bachet's Diophantos, though these are not the numbers in Metrodorus.

Simpson. Algebra. 1745. Section XI (misprinted IX in 1790), prob. XXIII, p. 93 (not in 1790 ed.). A says to B, if I had a of your money, I'd have as much as you together with half of C, etc., giving: x + a = y-a + ½z, y + b = z-b + ⅓x, z + c = x-c + ¼y. Finds general solution and does case a = b = c = 5.

Walkingame. Tutor's Assistant. 1751. 1777: p. 175, prob. 98; 1860: p. 184, prob. 97. (4, 1;  4, 2), flocks of sheep.

Mair. 1765? P. 458, ex. 7. Two men with money, like ass & mule: (1, 5; 1, 1).

Euler. Algebra. 1770. I.IV.IV.612: Question 3, pp. 208 209. Ass and mule: (1, 2; 1, 3).

Vyse. Tutor's Guide. 1771? Prob. 11, 1793: p. 130; 1799: p. 138 & Key p. 183. (1, 1; 1, 2) (= Euclid).

Dodson. Math. Repository. 1775.

P. 8, Quest. XIX. (1, 1; 1, 2) (= Euclid).

P. 19, Quest. L. Find x/y such that (x+1)/y = 1/3, x/(y+1) = 1/4. [(x+1)/y = 1/n; x/(y+1) = 1/(n+1) has solution x = n+1, y = (n+1)² - 1. In general, (x+1)/y  =  a/b, x/(y+1) = c/d gives x = c(a+b)/(ad-bc), y = b(c+d)/(ad-bc). One would normally assume a/b > c/d. This really belongs in 7.R.1. Cf Wolff in 7.R.3 for a different phrasing of the same problem.]

P. 31, Quest. LXXVI. Find x/y such that (x+4)/(y+4) = 4/3, (x-4)/(y-4) = 3/2. [In general (x+A)/(y+A) = a/b, (x-A)/(y-A) = c/d has solution x = A (2ac bc ad)/(bc-ad), y = A (ad+bc-2bd)/(bc-ad). One would normally assume a/b < c/d. This really belongs in 7.X.]

P. 46, Quest. XCIX. w = (x+y+z)/2, x = (w+y+z)/3, y = (w+x+z)/4, x = 14 + z.

Eadon. Repository. 1794. P. 296, no. 8. (5, 1; 5, 3).

D. Adams. Scholar's Arithmetic. 1801. P. 209, no. 2. (1, 1; 1, 2).

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. Art. 121, p. 32. (10, 2; 10, 3).

Unger. Arithmetische Unterhaltungen. 1838. Pp. 117-119 & 256-257, nos. 443 450. (5½, 2;  6½, 5), (13½, 7; 16½, 3), (10, 4; 7½, 2), (10, 4; 12, 3), (20⅓, 3; 12⅔, 4), (9½, 4;  15, 2), (33½, 2; 16½, 3), (6, 5; 4 2/5, 8).

Philip Kelland. The Elements of Algebra. A. & C. Black, Edinburgh, et al., 1839. ??NX. P. 134: "A's money or debt is a times B's; if A lose £10 to B, it will be b times B's." (Also entered in 7.X.)

The New Sphinx. c1840. No. 46, pp. 24 & 122. Women with baskets of eggs: (1, 2; 1, 1).

Fireside Amusements. 1850: No. 2, pp. 101 & 180; 1890: No. 2, p. 96. = New Sphinx. c1840.

The Family Friend (1856) 376, Enigmas, Charades, &c. 176 Arithmetical Puzzle. Standard (1, 2; 1, 1) given in a four stanza poem involving two costermongers with barrows of apples. Signed G. M. F. G.

Thomas Grainger Hall. The Elements of Algebra: Chiefly Intended for Schools, and the Junior Classes in Colleges. Second Edition: Altered and Enlarged. John W. Parker, London, 1846. P. 132, ex. 7. Appears to be (50, 2; 50, 1), but it reads: "A says to B, if you give me £50, I shall have twice as much as you had; but if I give you £50, each will have the same sum." The use of 'had' means the first equation is x + 50 = 2y, while the second equation is the usual x - 50 = y + 50. Answer: 250, 150.

Magician's Own Book. 1857. The two drovers, p. 246. (1, 1; 1, 2). = Book of 500 Puzzles, 1859, p. 60. = Boy's Own Conjuring Book, 1860, p. 218.

Vinot. 1860. Art. XLVI: L'Anesse et le Mulet, pp. 64-65. Gives a French translation of the Latin verse. (1, 1; 1, 2).

Leske. Illustriertes Spielbuch für Mädchen. 1864? Prob. 564-33, pp. 255 & 396. (1, 1; 1, 2). Notes that the solution to (a, 1; a, 2) is just a times the solution of the original.

(Beeton's) Boy's Own Magazine 3:4 (Apr 1889) 175 & 3:6 (Jun 1889) 255. (This is undoubtedly reprinted from Boy's Own magazine 1 (1863).) Mathematical question 34. (5, 1;  10, 2) with postage stamps.

Mittenzwey. 1880.