7. arithmetic & number theoretic recreations a. Fibonacci numbers

No. 123, p. 56. Same with horse worth 204, I (½, ⅓, ¼). Cf Gherardi 36-37.

No. 126, p. 58. 4 men buy a horse worth 37, I (1/2, 1/3, 1/4, 1/5). Answer:  1, 19, 25, 28. Cf Fibonacci 245-248.

No. 124 & 125, pp. 56 58. Complex formulations leading to the same problem with horse worth 2701 and 14800. No. 125 has the order of the constants reversed.

No. 140, pp. 60 61. 7 men buy a horse worth 100, II (2/3, 3/4, ..., 8/9). Answer:  (27630, 27855, 24460, 27175, 22830, 27265, 21640)/462.

No. 143, pp. 61 62. 4 men buy a horse worth 100 leading to II (1/3, 1/5, 1/6, 1/8) except the last man borrows from all the others. Answer:  (9747, 11058, 11875, 9348)/707.

Apianus. Kauffmanss Rechnung. 1527. Ff. M.iii.r - M.iii.v. x + y/3  =  y + z/2  =  z + 2y/3  = 30. Answer: (45, 45, 60)/2. Because x = y, this is actually the same as II-(1/3, 1/2, 2/3).

Georg von Peurbach. Elementa arithmetices, algorithmus .... Joseph Klug, Wittenberg, 1534, ??NYS. (There were several previous editions back to 1492, with variant titles. Rara 53 54. Glaisher, op. cit. in 7.G.1 under Widman, describes this extensively and gives the following on p. 97. This edition is substantially better than previous ones, but Peurbach died in 1461!)

F. D.iii.verso. I-(½, ⅓) with horse worth 10. Cf della Francesca 16r.

F. E.i.verso. II-(½, ⅓, ¼) with horse worth 100. Cf Lucca 1754 58r.

Cardan. Practica Arithmetice. 1539. Chap. 66, section 98, ff. HH.vi.r - HH.vi.v (p. 169). Men find a purse and buy a horse, giving: x + y + p/2  =  y + z + p/5  =  x + z + p/3  =  h. Answer:  6, 10, 15; 30, 31. Cf Lucca 1754 61r-61v; della Francesca 40v; Pacioli 193v-194r.

Tartaglia. General Trattato. 1556. Book 17, art. 8, 13, 14, 22, 32, 40, 41, pp. 268v, 269v, 271v, 273v, 275v-277r.

Art. 8. I-(⅓, ¼) with h_i = 14, 17. Answer:  (100, 162)/11.

Art. 13. II-(½, ⅓, ¼) with h_i = 40, 52, 62. Answer:  (584, 832, 1404)/25.

Art. 14. II-(2/3, 5/8, 4/5, 7/10) with house worth 200. Answer:  (2500, 3150, 2320, 2850)/23.

Art. 22. II-(½, ⅔, ¾). Answer: 48, 48, 36; 72, which is not in lowest terms. Cf AR 180.

Art. 32. I-(⅓, ¼) with h_i = 32, 42. Answer: (216, 408)/11.

Art. 40. I-(½, ⅓, ¼) with h = 40. Answer:  (200, 440, 520)/17. Cf Lucca 1754 58r.

Art. 41. I-(½, ⅓, ¼) with h = 20. Takes about two pages to finally get half of the preceding answer. Cf Lucca 1754 58r.

Buteo. Logistica. 1559.

Pp. 189-190. I-(½, ⅓) with h_i = 30, 20. Answer: 24, 12.

Pp. 190-192. I-(1/3, 1/4, 1/5) with h_i = 14, 8, 8. I find this remarkable in that he uses three unknowns -- A, B, C -- and solves by systematic elimination. Answer:  11, 4, 5.

Pp. 192-193. I-(1/3, 1/4, 1/5). He assumes the amount is 17 and gets: 5, 11, 13; 17. Cf Fibonacci 229.

Pp. 193-196. I-(1/2, 1/3, 1/4, 1/6) with h_i = 17, 12, 13, 13. Answer:  6, 4, 8, 10.

Prob. 81, pp. 289-291. Problem with soldiers, equivalent to I (1/2, 1/3, 1/4) with amount 14280. Answer:  4200, 9240, 10920. Cf Gherardi 36-37.

Prob. 30, pp. 357-358. Amounts desired from others are variable, giving x + y/2 + z/3  = 14, x/3 + y + z/4 = 13, x/6 + y/8 + z  =  14. Answer:  6, 8, 12.

Schott. 1674.

Ænigma V, p. 555. Two men want to buy a field worth 100, giving x + y/2 + 5 = 100 = y + x/3. Equivalent to I-(1/2, 1/3) with h_i = 95, 100. Answer: 54, 82.

Ænigma I, pp. 562-563. I-(1/3, 1/4) with h = 110. Answer: 80, 90. Cf al-Karkhi I 42.

Ænigma III, p. 563. I-(1/3, 1/4, 1/5) with h = 100. Answer: 52, 68, 76. Cf Diophantos 24.

"A Lover of the Mathematics." A Mathematical Miscellany in Four Parts. 2nd ed., S. Fuller, Dublin, 1735. Part III, no. 45, p. 94. Men buying a house worth 1200. I-(⅔, ¾). Answer: 800, 600. Cf Riese: Rechnung.

Simpson. Algebra. 1745. Section XI (misprinted IX in 1790), prob. XXV, pp. 95-96 (1790: prob. 34, pp. 94 95). II (1/2, 1/3, 1/4, 1/5) with h_i = a, b, c, d. Does example with values 357, 476, 595, 714 and answer:  190, 334, 426, 676. 1745 gives a second method.

Les Amusemens. 1749. Prob. 172, pp. 319-320. I (½, ⅓, ¼) with h = 40. Solves with a general h. Cf Gherardi 36-37.

Euler. Algebra. 1770. I.IV.

III: Question for practice, no. 19, p. 205. I (½, ⅓, ¼) with horse worth 34. Cf Gherardi 36-37.

IV.618: Question 5, pp. 212 213. I (⅔, ¾) paying a debt of 29. Cf Riese: Rechnung.

IV.619 620: Question 6, pp. 213 214. II (½, ⅓, ¼) buying a vineyard worth 100. Cf Lucca 1754 58r.

IV.621, pp. 214 215. Gives general form of solution of type II problem, using 4 person case as an example.

IV.622: Question 7, pp. 215 216. Problem equivalent to I (½, ⅓, ¼) with h = 901, i.e. 53/2 times the problem on p. 205. Cf Gherardi 36-37.

Vyse. Tutor's Guide. 1771? Prob. 13, 1793: p. 131; 1799: p. 138 & Key pp. 183-184. I (⅔, ¾) buying a horse worth 1200. Cf Riese: Rechnung.

Silvestre François Lacroix. Élémens d'Algèbre, a l'Usage de l'École Centrale des Quatre-Nations. 14th ed., Bachelier, Paris, 1825. Section 82, ex. 8, pp. 122-123. Variant version -- he has I-(1/2, 1/3) with h = 50,000 and then says the third could buy if he had 1/4 of the first's money. I think this is a corruption of I-(1/2, 1/3, 1/4). Cf della Francesca 16r.

Pearson. 1907. Part II, no. 139: The money boxes, pp. 141 & 218. I (1/2, 1/3, 1/4, 1/5) with h = 740. Cf Fibonacci 245-248.

M. Adams. Puzzles That Everyone Can Do. 1931. Prob. 101, pp. 43 & 144: A garaging problem. A variation on type IV. a + b + c + d + e = 100; a + b = 52; b + c = 43; c + d = 34; d + e = 30.

7.R.3. SISTERS AND BROTHERS
New section.

NOTATION: (a, b) means each boy has a times as many sisters as brothers, while each girl has b times as many brothers as sisters. This only has integer solutions for the integer pairs:

(a, b) = (1, 2), (1, 3), (2, 1), (2, 2), (3, 1), with solutions:

(Boys, Girls) = (4, 3), (3, 2), (3, 4), (2, 2), (2, 3).

See Wolff for a related problem.

See Fireside Amusements; Cutler for a related trick problem.

Fireside Amusements. 1850: No. 47, pp. 114 & 181; 1890: No. 33, p. 102. "Mr. Jones told another gentleman that he had six daughters, and each daughter had a brother; how many children had Mr. Jones?" Cf Cutler, below.

Mittenzwey. 1880. Prob. 15, pp. 2-3 & 59; 1895?: 17, pp. 8-9 & 63; 1917: 15, pp. 8 & 57. Sisters and brothers. (2, 1).

Peano. Giochi. 1924. Prob. 55, p. 15. (1, 2).

King. Best 100. 1927. No. 39, pp. 19 & 47. = Foulsham's, no. 11, pp. 7 & 11. (2, 1).

Heinrich Voggenreiter. Deutsches Spielbuch Sechster Teil: Heimspiele. Ludwig Voggenreiter, Potsdam, 1930. P. 108. (2, 1).

Dr. Th. Wolff. Die lächelnde Sphinx. Academia Verlagsbuchhandlung, Prague, 1937. Prob. 28, pp. 193 & 203. A man in an office says he has four times as many male colleagues as female colleagues. One of the women answers that she has five times as many male colleagues as female colleagues. This is (1/4, 5) in the above notation. In general, this would give m - 1 = af, m = b (f - 1), which has solution f = (b + 1)/(b   a). See Dodson in 7.R for a different phrasing of the same problem.

Depew. Cokesbury Game Book. 1939. Sisters and brothers, p. 218. (2, 1).

Owen Grant. Popular Party Games. Universal, London, nd [1940s?]. Prob. 8, pp. 36 & 50. (2, 1).

John Henry Cutler. Dr. Quizzler's Mind Teasers. Greenberg, NY, 1944. ??NYS -- excerpted in: Dr. Quizzler's mind teasers; Games Magazine 16:3 (No. 109) (Jun 1992) 47 & 43, prob. 27. "Mr. and Mrs. Twichell have six daughters. Each of the daughters has a brother. How many persons are there in the entire family?" Only 9 (counting the parents). Cf Fireside Amusements, above.

Joseph Leeming. Riddles, Riddles, Riddles. Franklin Watts, 1953; Fawcett Gold Medal, 1967. P. 113, no. 40. (1, 2).

Ripley's Puzzles and Games. 1966. P. 12. Man has two boys and a girl. He wants to have 12 boys and for each boy to have a sister. How many girls are needed?

Liz Allen. Brain Sharpeners. Op. cit. in 5.B. 1991. The sum of the siblings, pp. 31 & 110. (2, 1).

7.R.4. "IF I SOLD YOUR EGGS AT MY PRICE, I'D GET ...."
New section.
Simpson. Algebra. 1745. Section XI (misprinted IX in 1790), prob. XL, pp. 106-107 (1790: prob. LIII, p. 108). Market women bring x and y eggs to market, with x + y = 100 [c] and sell at prices A and B such that they get the same, i.e. Ax = By. First says to the second: "Had I brought as many eggs as you I should have received 18 [a] Pence for them". The other responds: "Had I brought no more than you, I should have received only 8 [b] Pence for mine". I.e. Ay = a, Bx = b. This gives (x/y)² = b/a, x  =  cb/[a+b] = bc/[b+ab], y = ca/[a+b].

Mittenzwey. 1880. Prob. 123, pp. 25 & 76; 1895?: 141, pp. 29 & 79; 1917: 141, pp. 27 & 77. Same as Simpson, with it being clearly phrased "If I had had your eggs and sold them at my price, ...", with c = 110, a = 250, b = 360. Answer is given as 50, 60, but ought to be as 60, 50. Here the prices also come out integral.

McKay. At Home Tonight. 1940. Prob. 11: Buying cows, pp. 64 & 78. Farmers A and B each buy £350 worth of cows. If A had bought at B's price, he would have paid £250. What would B have paid if he bought at A's price? Letting x, y be the numbers of cows and A, B be the prices, we have Ax = By = 350, Bx = 250, and we want Ay. Then y/x = 7/5 and Ay = Ax(7/5) = 490.
7.S. DILUTION AND MIXING PROBLEMS
See Tropfke 569.

There are a number of problems of this sort. One type is the same as the Hundred Fowls problem (7.P.1) where the solutions need not be integers. Lucca 1754, c1330, has a number of these. Here I consider only some of special interest and the following.

Recorde. Second Part. 1552. H&S quotes from the 1579 ed., f. Y.3. "It hath great use in composition of medicines, and also in myxtures of metalles, and some use it hath in myxtures of wines. but I wshe [sic] it were lesse used therein than it is now a daies." The 1668 ed., p. 295: The Rule of Mixture, has: "And it hath great use in composition of Medicines, and also in mixtures of Metalls, and some use it hath in mixtures of Wines : but I wish it were less used therein then it is now-a-days."
7.S.1. DISHONEST BUTLER DRINKING SOME AND REPLACING WITH

WATER
Dodson, Todhunter and Clark are problems to determine the amount taken off each time.
Papyrus Rhind, op. cit. in 7.C. Prob. 71, p. 108 of vol. 1 (1927) (= p. 57 of 1978 ed.). ¼ is poured off & replaced, what is the strength? (H&S 85 quotes Peet's version.)

Bakhshali MS. c7C. Kaye I 48; III 201-202, ff. 12r-12v. Man has bottle holding 4 prasthas of wine. (Drinks ¼ and refills with water) four times. How much wine is left?

Cardan. Practica Arithmetice. 1539. Chap. 66, sections 36 & 37, ff. DD.v.r - DD.v.v (p. 146). Drink three pitchers and replace with water four times leaving wine of half strength. Then the same for three times.

Tartaglia. Quesiti, et Inventioni Diverse, 1546, op. cit. in 7.E.1, Book 9, quest. 18, pp. 102v 103r. (Remove 2 and replace) thrice to halve strength.

Buteo. Logistica. 1559. Prob. 85, pp. 296-298. Butler drinking some and replacing with water. (7/8)⁵. (H&S 85)

Trenchant. Op. cit. in 7.L, 1566. 1578 ed., p. 297. ??NYS. (Remove  1/12th and replace) six times. (H&S 85 gives French and English. Sanford 209 gives English.)

Les Amusemens. 1749. Prob. 176, pp. 326-327. Sommelier drinks 6 pints from a cask of 360 and replaces with water three times -- how much wine has he drunk?

Dodson. Math. Repository. 1775. P. 76, Quest. CXLI. Cask of 81 gallons. (x is drawn off and replaced by water) four times, leaving 16 gallons of wine in the mixture. Gives a general solution.

Ozanam Montucla. 1778. Prob. 21, 1778: 212-214; 1803: 207-209. Prob. 20, 1814: 179 181; 1840: 93. Dishonest butler (removes 1/100th and replaces) 30 times. Notes that it is easier to use logarithms.

Bullen. Op. cit. in 7.G.1. 1789. Chap. 38, prob. 33, p. 243. Cask of 500 gallons; (remove 1/10th and replace with water) five times.

Bourdon. Algèbre. 7th ed., 1834. Art. 222, last problem, p. 364. Barrel of 100 pints of wine. One pint is drunk and replaced by water each day. How much wine is left after 50 days? When is the wine diluted to half its strength? One third? One quarter?

Anonymous. A Treatise on Arithmetic in Theory and Practice: for the use of The Irish National Schools. 3rd ed., 1850. Op. cit. in 7.H. P. 358, no. 35. Same as Bullen.

Vinot. 1860. Art. LVI: Le Sommelier infidèle, pp. 73-74. Barrel of 100. Sommelier drinks 1 and replaces with water 30 times. Computes the amount of water in the barrel each day.

Todhunter. Algebra, 5th ed. 1870. Examples XXXIII, no. 8, pp. 285 & 590. From 256 gallons of wine, draw off x and replace with water, four times to leave only 81 gallons in the container.

M. Ph. André. Éléments d'Arithmétique (N^o 3) a l'usage de toutes les institutions .... 3rd ed., Librairie Classique de F.-E. André-Guédon, Paris, 1876. Prob. 98, p. 62. Barrel of wine holding 210. Remove 45 and replace with water three times. Determine the amount of wine and water.

Mittenzwey. 1880.