7. arithmetic & number theoretic recreations a. Fibonacci numbers

Prob. 76, pp. 15 & 66; 1895?: 83, pp. 19 & 69; 1917: 83, pp. 18 & 65. Cask of 192 l worth 4 per l. Remove a quarter and replace with water. What is the value of the result? Repeat the dilution. What is the value now and how much less is the cask worth than originally?

Prob. 105, pp. 21-22 & 73; 1895?: 122, pp. 26 & 75. (In 1917, this was replaced.) Remove 4 from a cask and replace with water, thrice. The mixture now contains 2½ more water than wine. How big is the cask? Letting C be the size of the cask, this leads to [(C-4)/C]³ = (C/2 - 5/4)/C which is a cubic with one real root, C = 16. I am not surprised that this was dropped.

Charles Pendlebury & W. S. Beard. A "Shilling" Arithmetic. Bell, London, 1899 -- my copy is 59th ptg, 1960 and says it was reset for the 49th ptg of 1944. Examination Papers XIX, prob. 15, p. 192. A vessel is 17% spirit. When 10 has been drawn off and replaced with water, it is now 15 1/9 % spirit. How big is the vessel?

Dudeney. Weekly Dispatch (8 Feb 1903) 13. (Remove 1/100th and replace) 30 times.

Clark. Mental Nuts. 1904, no. 39. Find capacity of the keg. (Fill a keg from a 20 gallon cask and then replace with water) three times to dilute the cask to half-strength. How big is the keg?

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. A predecessor of this problem -- prob. 92, p. 61. Vase A has 12 litres of wine and 4 litres of water. Vase B has 8 litres of wine and 3 litres of water. Take off 4 litres from each and then put the 4 litres from A into B and the four litres from B into A. Determine contents in each vase.

Viscount John Allsebrook Simon. [Memory of Lewis Carroll.] IN: Appendix A of Derek Hudson; Lewis Carroll, An Illustrated Biography; Constable, (1954); new ed., 1976, pp. 248 249. = Carroll-Wakeling II, prob. 34: Two tumblers, pp. 52 & 73. 50  spoonfuls of brandy and 50 spoonfuls of water -- transfer a spoonful back and forth. He says Carroll posed this. Mentioned in Carroll-Gardner, p. 80, who gives the full name. The DNB says he entered Wadham College, Oxford, in 1892, and his Memory says he met Carroll then. So this dates from 1892, but Carroll could have been propounding it years before.

Ball. MRE, 3rd ed., 1896, pp. 26 27. Water in wine versus wine in water. He says this is a question "which I have often propounded in past years". Not in the 1st ed of 1892.

W. P. Workman. The Tutorial Arithmetic, op. cit. in 7.H.1. 1902. Section IX, examples CXLV, prob. 35, pp. 427 & 544 (= 433 & 577 in c1928 ed.). A contains 11  pt water and 7 pt wine, B contains 5 pt water and 13 pt wine. Transfer 2 pt from A to B and back. Find changes of water and wine in both A & B. This is a precursor of the puzzle idea.

Pearson. 1907. Part II, no. 18, pp. 117 & 194 195. Butter in lard versus lard in butter.

Loyd. Cyclopedia, 1914, pp. 287 & 378. Forty quarts milk and forty quarts water with a quart poured back and forth. He says the ratios of milk to water are then 1 : 40 and 40 : 1, which is correct, but isn't the usual question.

H. E. Licks. Op. cit. in 5.A. 1917. Art. 20, pp. 16 17. Water and wine.

Ahrens. A&N, 1918, p. 89.

T. O'Conor Sloane. Rapid Arithmetic. Quick and Special Methods in Arithmetical Calculation Together with a Collection of Puzzles and Curiosities of Numbers. Van Nostrand, 1922. [Combined into: T. O'Conor Sloane, J. E. Thompson & H. E. Licks; Speed and Fun With Figures; Van Nostrand, 1939.] Wine and water paradox, pp. 168 169.

F. & V. Meynell. The Week End Book. Op. cit. in 7.E. 1924. Prob. two: 2nd ed., p. 274; 5th?? ed., p. 407. Find "proportion of the amount of water in A to the amount of milk in B."

Peano. Giochi. 1924. Prob. 24, p. 7. Water and wine.

Loyd Jr. SLAHP. 1928. Cheating the babies, pp. 40 & 98. Two large cans with 10 gallons of milk and 10 of water. Pour 3 gallons back and forth. "Have I more milk in the water can than I have water in the milk can?" He works out that each can has the proportion 7 9/13 : 2 4/13 [= 100 : 30 = 10 : 3].

Phillips. Week End. 1932. Time tests of intelligence, no. 25, pp. 17 & 190. Whisky and water in equal amounts. He asks about proportions rather than amounts.

Phillips. The Playtime Omnibus. Op. cit. in 6.AF. 1933. Section XVII, prob. 17: The two vessels, pp. 56 & 238. Same as in his Week End.

Abraham. 1933. Prob. 26 -- Whiskey and water, pp. 10 & 24 (7 & 112).

Perelman. FMP. c1935? Water and wine, pp. 215 & 218.

Phillips. Brush. 1936. Prob. D.5: Whisky and water, pp. 12 & 82. Same as in his Week End.

Dr. Th. Wolff. Die lächelnde Sphinx. Academia Verlagsbuchhandlung, Prague, 1937. Prob. 36, pp. 194 & 204. Man takes a cup of black coffee. He drinks 1/6 of it and fills it up with milk. He then drinks 1/3 of it and fills it up with milk. He then drinks half of it and fills it up with milk. Then he drinks the whole cup. Has he drunk more milk than coffee or vice-versa?

McKay. Party Night. 1940. No. 6, pp. 176-177. Water and wine. "I have seen, after dinner, parties discuss the problem for a long time, ...."

Sullivan. Unusual. 1943. Prob. 21: Mixtures. Gasoline and alcohol.

E. P. Northrop. Riddles in Mathematics. 1944. 1944: 14-16; 1945: 13 15; 1961: 23 25. Water and milk.

Gamow & Stern. 1958. Gin and tonic. Pp. 103 104.
7.T. FOUR NUMBER GAME
The game takes (a, b, c, d) to (a   b, b - c, c   d, d   a). See 7.BB for other iterated functions.
C. Ciamberlini & A. Marengoni. Su una interessante curiosità numerica. Period. Mat. (4) 17 (1937) 25 30. They attribute the problem to E. Ducci.

G [= J. Ginsburg]. Curiosa 30. An interesting observation. SM 5 (1938) 135. Brief report on above article.

Benedict Freedman. The four number game. SM 14 (1948) 35 47. No references. Obtains basic results for n number version.

S. P. Mohanty. On a problem of S. J. Bezuszka & M. J. Kenney on cyclic difference of pairs of integers. Fibonacci Quarterly 19 (1981) 314 318. (This and three of its references are not in Meyers.)

Leroy F. Meyers. Ducci's four number problem: a short bibliography. CM 8 (1982) 262 266. See Ludington-Young, below, for nine additional references.

M. Gardner. Riddles of the Sphinx and Other Mathematical Puzzle Tales. New Math. Library, MAA, 1987. Prob. 29: Hustle off to Buffalo, parts 2 5, pp. 134 136, 151 152, 160 163. Gives a proof that most quadruples converge to all zeroes and finds the quadruples that cycle.

Joseph W. Creely. The length of a three-number game. Fibonacci Quarterly 26:2 (May 1988) 141-143. Solves three and two number versions.

Stanley Bezuszka with Lou D'Angelo & Margaret J. Kenny. The Wonder Square. Booklet 2, Boston College Math. Inst. Motivated Math Project Activity. Boston College Press, Chestnut Hill, Mass., 1976. 32pp. Studies the process with various special numbers, e.g. progressions, Fibonacci, Tribonacci and figurate numbers. The Tribonacci case produces starting sequences with length n for any n.

Anne Ludington-Young. The length of the n-number game. Fibonacci Quarterly 28:3 (Aug 1990) 259-265. Obtains a bound and solves some cases. Cites Meyers and 9 additional references.
7.U. POSTAGE STAMP PROBLEM
What integers are non negative integral combinations of a, b, ...? In particular, what is the largest integer which is not such a combination? This is well known for the case of two values, but remains unknown for more values. From about 1960 onward, the case with two values frequently occurs in number theory texts and as a puzzle problem, but I haven't entered such appearances.
Dickson, vol. II, chap. II is uncharacteristically obscure about this. It is generally attributed to Frobenius (1849-1917).

J. J. Sylvester. Math. Quest. Educ. Times 41 (1884) 21. ??NYS. Solves the problem for two values.

Alfred Brauer. On a problem of partitions. Amer. J. Math. 64 (1942) 299 312. On p. 301, he says that some of the results are due to I. Schur in a lecture in 1935, others are due to himself and others are joint work. He cites Sylvester, but doesn't mention Frobenius.

Alfred Brauer & B. M. Seelbinder. On a problem of partitions II. Amer. J. Math. 76 (1954) 343 346. "A number of years ago, the first of the authors studied together with I. Schur the following problem of Frobenius: ..."

Alfred Brauer & James E. Shockley. On a problem of Frobenius. J. reine angew. Math. 210 (1962) 215 220. "G. Frobenius, in his lectures, raised the following problem repeatedly .... No result was obtained for many years. In 1935, I. Schur proved in his last lecture in Berlin the following result ...."

Ernest S. Selmer. On the linear diophantine problem of Frobenius. J. reine angew. Math. 293/294 (1977) 1 17. Gives 25 references which he believes to be a complete list. Cites Sylvester, but the next oldest are the 1942 and 1954 papers above. The 1962 paper above is the first to mention Frobenius in the title.

Goldbach. Letter to D. Bernoulli, 31 Jan 1729. Ibid., pp. 280 283. Reply to the above. Setting y = ax, he gives an easy proof for the only integer solutions. He says the fractional solutions are (f/g)^g/(f g).

L. Euler. Introductio in Analysin Infinitorum. Bousquet, Lausanne, 1748. Vol. 2, § 519, pp. 295-296 & Tab. XXV, fig. 103. = Introduction to the Analysis of the Infinite; trans. by John D. Blanton; Springer, NY, 1988-1990; Book II, pp. 339-340 & 489. Gets x = (1 + 1/n)ⁿ; y = (1 + 1/n)ⁿ⁺¹.

L. Euler. De formulis exponentialibus replicatus. Acta Acad. Sci. Petropol. 1 (1777(1778)) 38 60. = Opera Omnia (1) 15 (1927) 268 297. Iterated exponentials.

Ahrens. A&N, 1918, pp. 76 78, discusses the problem and notes that Goldbach's fractional solution is rational if and only if g/(f g) is integral, say n, which gives x  =  (1 + 1/n)ⁿ.

C. A. B. S[mith]. 5-minute problem. Eureka 3 (Jan 1940) 4 & 24. Letting p, m, c be the number of physicists, chemists and mathematicians at a lecture, he observes that p^m = m^c, c^m = m^p, c^p = p^c and m > c. One finds p = c, hence c^m = m^c.

R. L. Goodstein. Note 1725: The equation a^b = b^a. MG 28 (No. 279) (May 1944) 76. Quick derivation of general solution and the rational solutions.

Anonymous. The problems drive. Eureka 14 (Oct 1951) 12-13 & 22. No. 5. Find rational solutions of p^q = q^p.

R. A. Knoebel. Exponentials reiterated. AMM 88 (1981) 235 252. Extensive history and bibliography.

F. Gerrish. Note 76.25: a^b = b^a: the positive integer solution. MG 76 (No. 477) (Nov 1992) 403. Short note on the integer case with two recent references.

R. F. Churchhouse. Solutions of the equation x^y = y^x. Bull. Inst. Math. Appl. 31:7/8 (Jul/Aug 1995) 106. Easily finds the real solutions, then the rational and integer solutions. Notes that x = i, y = -i is a solution!
7.W. CARD PILING OVER A CLIFF
Identical cards (or dominoes) of length 1 can be stacked to reach out from the edge of a cliff. The simplest analysis shows that n cards can reach out 1/2 + 1/4 + 1/6 + ... + 1/2n    (log n)/2. Some authors consider real dominoes which can be piled in several orientations.
J. G. Coffin, proposer. Problem 3009. AMM 30 (1923) 76. Asks for maximum overhang for n cards. (Never solved!)

Max Black. Reported in: J. F. O'Donovan; Clear thinking; Eureka 1 (Jan 1939) 15 & 20. Problem 1. Asks for maximum extension from the bottom card with 52 cards, then for n cards. Solution is 1/2 + 1/4 + 1/6 + ... + 1/(2n-2). For n = 52, this is about 9/4.

A. S. Ramsey. Statics. 2nd ed., CUP, 1941, example 4.68, pp. 47 48. Discusses equal spacing with a support at the outer end, e.g. a staircase. (Is this in the 1st ed. of 1934?? My source indicated that Ramsey cited a Tripos exam.)

Heinrich Dörrie. Mathematische Miniaturen. Ferdinand Hirt, Breslau, 1943; facsimile reprint by Martin Sändig, Wiesbaden, 1979. Prob. 240, pp. 279-282. Using cards of length 2, he gets the extension 1/1 + 1/2 + 1/3 + ... and shows the curve formed is x = log (y/t), where t is the thickness of the cards and x and y are measured from the outer end of the top card -- in the usual picture, they are both going negatively.

R. T. Sharp, proposer; C. W. Trigg, solver. Problem 52. Pi Mu Epsilon J. ?? & (April 1954) 411 412. Shows overhang approaches infinity, but the proposal asks for the largest overhang for n dominoes, which is not answered. Notes that the dominoes can be angled so the diagonal is perpendicular to the cliff edge. This is also in Trigg; op. cit. in 5.Q; Quickie 52: Piled dominoes, pp. 17 & 99, but it still doesn't answer the proposal.

P. B. Johnson, proposer; Michael Goldberg, Albert Wilansky, solvers. Problem E1122 -- Stacking cards. AMM 61 (1954) 423 & 62 (1955) 123 124. Both show overhang can go to infinity.

Paul B. Johnson. Leaning tower of lire. Amer. J. Physics 23 (Apr 1955) 240. Claims harmonic series gives greatest overhang!!

P. J. Clarke. Note 2622: Statical absurdity. MG 40 (No. 333) (Oct 1956) 213 215. Considers homogeneous bricks with weight bounded above and below and length bounded below. Then one can take such bricks in any order to achieve an arbitrarily large overhang.

Gamow & Stern. 1958. Building blocks. Pp. 90 93.

D. St.P. Barnard. Problem in The Observer, 1962. ??NYS.

D. St.P. Barnard. Adventures in Mathematics. Chap. 8 -- The Domino Story. (1965); Funk & Wagnalls, NY, 1968, pp. 109 122. Gets 3.969 for 13 dominoes.

Birtwistle. Math. Puzzles & Perplexities. 1971. Overbalance, pp. 122-124. Standard piling. Shows the harmonic series diverges.

See A. K. Austin, 1972, in 5.N for a connection with this problem.

Stephen Ainley. Letter: Finely balanced. MG 63 (No. 426) (1979) 272. Gets 1.1679 for 4 cards.

[S. N. Collings]. Puzzle no. 47. Bull. Inst. Math. Appl. 15 (1979) 268 & 312. Shows that a simple counterbalancing scheme gets m/2 for 2^m   1 dominoes, so the overhang for n dominoes is at least ½ log₂(n+1).

R. E. Scraton. Letter: A giant leap. MG 64 (No. 429) (1980) 202 203. Discusses some history, especially Barnard's problem.

Nick Lord, proposer; uncertain solver. Problem 71.E. MG 71 (No. 457) (Oct 1987) 236 & 72 (No. 459) (Mar 1988) 54 55. Overhang can diverge even if the lengths converge to zero.

Jeremy Humphries, proposer; various solvers. Prob. 129.5 -- Planks. M500 129 (Oct 1992) 18 & 131 (Feb 1993) 18. Uniform planks of lengths 2, 3, 4. Find maximum overhang. Best answer is 3 5/18.

Unnamed solver, probably Jeremy Humphries. Prob. 145.2 -- Overhanging dominoes. M500 145 (?? 1995) ?? & 146 (Sep 1995) 17. He uses approximately real dominoes of size .7 x 2.2 x 4.4 cm. Given three dominoes, how far out can you reach? And how far away from the edge can one domino be? Answer is basically independent of the shape. Place one domino on its large face with its centre of gravity at the cliff edge and its diagonal going straightout. For maximum reach, similarly place another domino with its centre of gravity at the far corner of the first domino. To get a domino maximally far from the edge, place one on edge so its largest face is parallel to the cliff edge and with its centre of gravity at the far corner of the first domino. In both cases, counterbalance by putting the other domino with its centre of gravity at the inner corner of the first domino. So the answer to the first problem is the longest face diagonal and the answer to the second problem is half the longest face diagonal minus half the thickness.
7.X. HOW OLD IS ANN? AND OTHER AGE PROBLEMS
The simplest age problems are 'aha' problems, like 'Diophantos's age', going back at least to Metrodorus and will not be considered here -- see Tropfke 575 for these. More complicated, but still relatively straightforward age problems appear in various 16-19C arithmetic and algebra works, e.g. Schooten; Recorde; Baker; Cocker; Pike; American Tutor's Assistant; Ainsworth & Yeats, 1854, op. cit. in 7.H.4; Colenso. I have included only a sample of these to show the background and a few earlier versions. Simple problems of Form III then appear from the mid 18C and later in standard arithmetic books, and later in puzzle books like Hoffmann; Pearson; Loyd; Dudeney and Loyd Jr. These usually lead to two equations in two unknowns, a bit like 7.R, or one equation in one unknown, depending on how one sets up the algebra. In the 19C, this problem was popular in discussions of algebra as the problem can have a negative solution, which means that the second time is before the first rather than after, and so the problem was used in discussions of the existence and meaning of negative numbers -- see: Hutton, 1798?; Kelland; De Morgan (1831, 1836, 1840). About 1900, we get the "How old is Ann (or Mary)" versions, forms I and II below.

Form I: "The combined ages of Mary and Ann are 44 years, and Mary is twice as old as Ann was when Mary was half as old as Ann will be when Ann is three times as old as Mary was when Mary was three times as old as Ann. How old is Ann?" Answer is 16½; Mary is 27½. See: Pearson; Kinnaird; Loyd; Bain; White; Dudeney; Loyd Jr; Grant; Ransom; Doubleday - 2.

Form II: "Mary is 24. She is twice as old as Ann was when Mary was as old as Ann is now. How old is Ann?" Answer is 18. See: Haldeman-Julius; Menninger; Ransom; Doubleday - 2.

Other examples of this genre: Baker; Vinot; Brooks; Gibney; Lester; Clark; M. Adams; Meyer; Little Puzzle Book; Dunn.

Form III (a, b, c). X is now a times as old as Y; after b years, X  is c times as old as Y. I.e. X = aY, X + b = c (Y + b). This can be rephrased depending on the time of narration -- e.g. X is now c times as old as Y, but b years ago, X was only a times as old as Y. See Clark for an equivalent problem with candles burning.

If the problem refers to both some time ago and some time ahead, it is the same as a form usually stated about increasing or decreasing both the numerator and denominator of a fraction X/Y, i.e. (X-a)/(Y-a) = b; (X+c)/(Y+c) = d. See: Simpson; Dodson; Murray; Vinot. I have seen other examples of this but didn't note them.

Form IV-(a, b, c). X is now a; Y is now b; when will (or was) X be c times as old as Y? I.e. a + x = c (b + x).

General solution of Form III occurs in: Milne; Singmaster.

General solution of Form IV occurs in: De Morgan (1831).
Examples of type III.
a b c
10/7 2 4/3 Boy's Own Mag.

2 12 8/5 Milne

2 30 7/5 Dilworth

2 60 5/4 Dilworth

7/3 6 13/9 Milne

3 10 2 Vinot

3 10½ 2 Walkingame, 1751

3 14 2 Murray; Hummerston

3 15 2 Ladies' Diary; Mair; Vyse; Amer. Tutor's Asst.; Eadon; Bonnycastle, 1815; Child; Walkingame, 1835; New Sphinx; Charades, etc.; Pearson;

3 18 2 Dudeney

4 -2 6 Unger

4 4 3 Tate

4 5 3 Young

4 10 5/2 Tate

4 14 2 Simpson

5 5 3 Dudeney

5 5 4 Hutton, c1780?; Hutton, 1798?; Treatise;

6 3 4 Unger; Milne

6 24 2 Clark, 1897

7 3 4 Berkeley & Rowland

7 15 2 Unger

9 3 3 M. Adams

14 10 5⅓ Unger
Examples of type IV.
a b c
a b n De Morgan, 1931?

a b 3 De Morgan, 1831?

a b 4 Bourdon

32 5 10 Perelman

40 8 3 De Morgan, 1831?

40 9 2 Young

40 20 3 Kelland; Carroll-Gardner

42 12 4 Hutton, 1798?

45 12 3 Hoffmann; Dudeney

45 15 4 Bourdon

48 12 3 Vinot

48 12 7 Vinot

50 40 2 De Morgan, 1840

54 18 2 Manuel des Sorciers

56 29 2 De Morgan, 1836

62 30 5 Colenso

71 34 2 Hoffmann

71 34 3 Hoffmann

See Young World for a variant form equivalent to

2 2 -20
Robert Recorde. The Whetstone of Witte. John Kyngstone, London, 1557. Facsimile by Da Capo Press, NY & Theatrum Orbis Terrarum, Amsterdam, 1969. A question of ages, ff. Gg.i.v - Gg.ii.r. (The gathering number at the bottom of folio ii is misprinted G.) Father and two sons. B = A + 2. C = A + B + 4. A + B + C = 96.

Baker. Well Spring of Sciences. 1562? Prob. 1, 1580?: ff. 189r 190r; 1646: pp. 297-299; 1670: pp. 340-341. A is 120. B says if he were twice his present age, he would be as much older than A as A is older than B is now. C says the same with three times his age, D with four and E with five.

Frans van Schooten Sr. MS algebra text, Groningen Univ. Library, Hs. 443, c1624, f. 54r. ??NYS -- reproduced, translated and described by Jan van Maanen; The 'double--meaning' method for dating mathematical texts; IN: Vestigia Mathematica; ed. by M. Folkerts & J. P. Hogendijk; Rodopi, Amsterdam, 1993, pp. 253--263. "A man, wife and child are together 96 years, that is to say the man and child together 2 years more than the wife and the wife with the child together 14 years more than the man. I ask for the years of each." Van Maanen feels that this problem probably describes the van Schooten family and it is consistent with the family situation for a period in 1623--1624, when the man was 41, the wife 47 and the child 8, thereby giving a reasonable date for the undated MS. Van Schooten's two sons also used the problem in their works, but reversed the role of man and wife because it was not common for the wife to be so much older.

Edward Cocker. Arithmetic. Op. cit. in 7.R. 1678. Several problems, of which the following are the most interesting.