7. arithmetic & number theoretic recreations a. Fibonacci numbers

The unfaithful knave, pp. 4-5. 32 wine bottles, 9 on a side, reduced to 28, 24, 20.

The blind abbot and his monks, pp. 5-6. 24 monks, 9 on a side, changing to 20, 28, 32, 18.

Leske. Illustriertes Spielbuch für Mädchen. 1864?

Prob. 585-4, pp. 296 & 409: Der Kasten des Juweilers. 32 rings reduced to 28.

Prob. 585-6, pp. 296 & 409: Des Müllers Säcke. 32 sacks reduced to 28.

Bachet-Labosne. Problemes. 3rd ed., 1874. Supp. prob. VI, 1884: 189-191. 60 bottles diminished to 44.

Kamp. Op. cit. in 5.B. 1877. No. 16, pp. 325 326. Nine wine bottles on each side on the square. Servant steals some.

Mittenzwey. 1880.

Prob. 224, pp. 41 & 92; 1895?: 249, pp. 45 & 94; 1917: 249, pp. 40-41 & 89. 56 coins with 23 on a side, but with all corners the same and all edges the same, which gives a unique solution.

Prob. 225, pp. 41 & 92; 1895?: 250, pp. 45 & 94; 1917: 250, pp. 41 & 89. 32 flasks of wine with 9 on a side, reduced to 30, 28, 24, 20.

1895?: prob. 251, pp. 45 & 94; 1917: 249, pp. 41-42 & 90. 40 flasks of champagne, 11 on a side, reduced to 36, 32, 28, 24, 22.

Prob. 229, pp. 421 & 93; 1895?: 256, pp. 46 & 94; 1917: 256, pp. 42-43 & 90. 32 sacks, 12 on a side, reduced to 28. = Leske, 585-6.

Prob. 230-232, pp. 42-43 & 93; 1895?: 257-259, pp. 46-47 & 94-95; 1917: 257-259, pp. 43 & 90. Abbess and 24 nuns, 9 on a side, changed to 20, 28, 32.

Prob. 233, pp. 43 & 93; 1895?: 260, pp. 47 & 95; 1917: 260, pp. 43 & 91. Fortress with 600 defenders, 200 on a side, changed to 250, then 300, on each side.

Cassell's. 1881. Pp. 98 99: The twenty four monks. = Manson, 1911, pp. 249-250.

Handy Book for Boys and Girls. Op. cit. in 6.F.3. 1892. Pp. 38-39: The counter puzzle. "In an old book published over half a century ago, I came across this puzzle ...." Rearranges 24 as 20, 28, 32.

Somerville Gibney. So simple! V. -- A batch of match tricks. The Boy's Own Paper 20 (No. 988) (18 Dec 1897) 188-189. 24 changed to 25, 20, 28, 32, 30.

Dudeney. Problem 70: The well and the eight villas -- No. 70: The eight villas. Tit Bits 33 (5 Mar 1898) 432 & 34 (2 Apr 1898) 8. How many ways can numbers be placed in the 8 cells to make 9 along each side? Answer is 2035. He gives a general formula.

H. D. Northrop. Popular Pastimes. 1901. No. 3: The blind abbot and the monks, pp. 66-67 & 72. = The Sociable.

Dudeney. The monk's puzzle. London Mag. 9 (No. 49) (Aug 1902) 89 91 & 9 (No. 50) (Sep 1902) 219. (= CP, prob. 17, pp. 39 40 & 172 173.) How many ways can numbers be placed in the 8 cells to make 10 along each side?

Benson. 1904. The dishonest servant puzzle, p. 228. 28 bottles, 9 on each edge, reduced to 24, then 20.

Wehman. New Book of 200 Puzzles. 1908.

Pp. 20-21: The blind abbot and the monks. = The Sociable.

P. 22: Fifteen "square" puzzle. Uses the unusual figure of Endless Amusement II, prob. 32. Starts with 5 marks in each cell so that it adds to 15 each way. Remove four marks. Solution is a bit unclear.

M. Adams. Indoor Games. 1912. The cook and the jam, pp. 353-354. 36 jars of jam.

Blyth. Match-Stick Magic. 1921. Escaping from Germany, pp. 75-76. 32 with 9 on a side -- arranged 1, 7, 1 -- reduced to 28, 25 and 24.

Will Blyth. Money Magic. C. Arthur Pearson, London, 1926. The stolen tarts, pp. 91-95. 9  on a side, start with 32, reduce to 28, 24, 20.

Rohrbough. Brain Resters and Testers. c1935. Ba Gwa, p. 18 (= pp. 18 19 of 1940s?). 3 x 3 frame. Two player game. Start with 7 in each corner and 1 in each edge. A player places a extra counter in the frame and the other tries to rearrange to preserve 15 in each outside row. It says you can get 56 men on the board. [I can get 60 if the corners can be empty.]

Jeffrey J. F. Robinson. Musings on a problem. MTg 37 (1966) 23 24. A farmer has 41 cows and wants to see 15 along each side of his house which is in the centre of a 3 x 3 array. How many solutions are there if only 1, 2, ..., 8 different values can be used? He finds all the solutions in some cases.

Ripley's Puzzles and Games. 1966. Pp. 64-65, item 1. 16 pigs in 8 pens with 6 along each side. Add four pigs.

J. A. Dixon & Class 3T. Number squares. MTg 57 (1971) 38 40. Use the digits 1   8 so the four side totals are the same. They find that the sum can only be: 12, 13, 14, 15, with 1, 2, 2, 1 solutions. They then use 8 digits from 1   9 and find 35 solutions.
7.Q.1. REARRANGEMENT ON A CROSS
The counts from the base to the top and to the end of each arm remain constant though some (usually 2) of the pearls or diamonds have been removed. Trick versions with doubling up are in 7.Q.2.

Versions with a T or Y: Secret Out; M. Adams;

Pacioli. De Viribus. c1500. Ff. 116r - 117v. Cap. LXX. D(e). un prete ch' in pegno la borscia del corporale con la croci de p(er)le al Giudeo (Of a priest who pledges to a Jew the burse of the corporale with a cross of pearls). 15 with three on each arm, one counts to nine from the base to each arm end. This is reduced to 13. Asks how one can add one pearl and produce a count of ten -- answer is to put it at the base.

Thomas Hyde. Historia Nerdiludii, hoc est dicere, Trunculorum; .... (= Vol. 2 of De Ludis Orientalibus, see 7.B for vol. 1.) From the Sheldonian Theatre (i.e. OUP), Oxford, 1694. De Ludo dicto Magister & Servus, pp. 234-236. Cross with 28 thalers with 16 from the base to the end of each arm. He points out that the picture has an error causing the count to the ends of the side arms to be 17. Discusses general solution. Says it is known to the Arabs of the Holy Land.

Les Amusemens. 1749. P. xxvi. 13 markers reduced to 11.

Manuel des Sorciers. 1825. Pp. 136-138, art. 16. ??NX Cross of coins reduced from 13 to 11.

Endless Amusement II. 1826? Prob. 11, pp. 195-196. 15 diamonds reduced to 13.

The Boy's Own Book. The curious cross. 1828: 414; 1828-2: 420; 1829 (US): 213; 1855: 568; 1868: 628. 13 markers, reducing to 11.

Nuts to Crack III (1834), no. 80. The curious cross. Almost identical to Boy's Own Book.

The Riddler. 1835. The curious cross, p. 6. Identical to Boy's Own Book.

Young Man's Book. 1839. P. 60. Easy Method of Purloining without Discovery. Identical to Endless Amusement II, except with a title.

The New Sphinx. c1840. The curious cross, p. 143. Same as Boy's Own Book, with a few words changed.

The Sociable. 1858. Prob. 25: The dishonest jeweller, pp. 295 & 310. 15 diamonds, reducing to 13. = Book of 500 Puzzles, 1859, prob. 25, pp. 13 & 28. = Wehman, New Book of 200 Puzzles, 1908, p. 7.

The Secret Out. 1859. The Dishonest Servant, pp. 78-80. T shape. 16 coins reduced to 14. Presentation of the problem is done with cards.

Boy's Own Conjuring Book. 1860. Easy method of purloining without discovery, p. 295. 15 diamonds, reducing to 13. Very similar to The Sociable.

Leske. Illustriertes Spielbuch für Mädchen. 1864? Prob. 585-5, pp. 296 & 409: Juwelenkreuz. 18 jewels reduced to 16.

Mittenzwey. 1880. Prob. 226, pp. 41-42 & 92; 1895?: 252-253, pp. 45-46 & 94; 1917: 252 253, pp. 42 & 90. 15 jewels reduced to 13. The 1895? addition has 10 reduced to 8.

Lucas. La croix de perles. RM2, 1883, pp. 134 135. c= Lucas; L'Arithmétique Amusante; 1895; pp. 10-11. 15 reduced to 13 and discussion.

Lemon. 1890. The puzzling pearls, no. 535, pp. 69 & 117. 15 reduced to 13.

Don Lemon. Everybody's Pocket Cyclopedia. Revised 8th ed., 1890. Op. cit. in 5.A. P. 136, no. 7. Diamond cross reduced from 15 to 13. No solution.

É. Ducret. Récréations Mathématiques. Op. cit. in 4.A.1. 1892? P. 104: Les croix de jetons. Rearrange a cross of 13 to 11.

H. D. Northrop. Popular Pastimes. 1901. No. 4: The dishonest jeweller, pp. 67 & 72. = The Sociable.

M. Adams. Indoor Games. 1912 The dishonest steward, pp. 22 & 24. Rearrangement of a Y.

Dudeney. AM. 1917. Prob. 423: The ruby brooch, pp. 144-145 & 249. Complicated version. Brooch is a circle with 8 radii and the owner used to count from the centre to the circle, along ⅛ of the circumference and then back in, getting 8 each time. Dishonest jeweller reduces stones from 45 to 41 in a symmetric pattern of one in the middle, two on each arm and three on each arc. What was it before?

Blyth. Match-Stick Magic. 1921. Counting the cross, p. 33. Rearrange a cross of 15 to 13.
7.Q.2. REARRANGE A CROSS OF SIX TO MAKE TWO LINES OF

FOUR, ETC.
For the standard version, one has to put one marker on top of the one at the crossing. See 6.AO and 7.Q for some similar trick versions. The one which is closest to this section is:

(12, 7, 4) -- Trick version of a 3 x 3 square with doubled diagonal: Hoffmann (1876), Mittenzwey, Hoffmann (1893), no. 8.

See also in 6.AO, Hoffmann (1893), no. 9.
Les Amusemens. 1749. P. xxv. A cross of six -- four crossing three -- rearranged to count four both ways.

Blyth. Match-Stick Magic. 1921. Five by count, pp. 33-34. A cross of seven -- five crossing three -- rearranged to count five both ways.

J. F. Orrin. Easy Magic for Evening Parties. Jarrolds, London, nd [1930s??]. The five puzzle, pp. 66-67. As in Blyth.

Sid G. Hedges. More Indoor and Community Games. Methuen, London, 1937. Penny puzzle, pp. 51-52. Four pennies in the shape of a T or Y tetromino. Make two rows of three. Put one from an end of the row of three on the crossing.

Depew. Cokesbury Game Book. 1939. Seven coins, p. 223. As in Blyth.

Ripley's Puzzles and Games. 1966. Pp. 18-19, item 2. An L with four in one leg and three in the other rearranged to have four on both lines.

7.R. "IF I HAD ONE FROM YOU, I'D HAVE TWICE YOU"
Jacques Sesiano. The appearance of negative solutions in mediaeval mathematics. Archive for the History of the Exact Sciences 32 (1985) 105-150. In this, he discusses problems of the types given here and in 7.P.1, 7.R.1, 7.R.2 and 7.P.7. He pays particular attention to whether the author discusses the problems in general or recognizes conditions for positivity or consistency, covering this in more detail than I do here.
See Tropfke 609.

In the medieval period, all these problems are extended in various ways to lead to quadratic and higher equations, but I think these become non-recreational although they certainly were not practical at the time, except for a few involving compound interest.

I include a few examples of unusual cases of two equations in two unknowns here.
NOTATION.

(a, b; c, d) denotes the general form for two people.

"If I had a from you, I'd have b times you."

"And if I had c from you, I'd have d times you."

I (a, b; c, d; ...) denotes the case for more people where 'you' means all the others.

II (a, b; c, d; ...) denotes the same where 'you' means just the next person, taken cyclically.

With two people, there is no need to distinguish these cases.

See Alcuin for an example where the second statement is interpreted as occurring after the first is carried out.

Sometimes a = 0 -- see: Kelland (1839); Hummerston (1924). If a = c = ... = 0, this can interpreted as a form of 7.R or 7.R.1 -- see: Dodson (1775).

See Ghaligai, Cardan for versions with "If I had a times yours from you, I'd have b times you", i.e. x + ay = b(y - ay), ....

Versions giving (x-a)/(y-a) = c; (x+b)/(y+b) = d, etc. are forms of Age Problems and are generally placed in 7.X. But see: Dodson.

See Hall for a version where the first equation is x + a = by.

There are versions where one asks for a fraction x/y such that (x+a)/y = c/d, x/(y+b) = e/f. These are forms of 7.R.1. See: Dodson; Hall.

Let T be the total of the amounts. Then I (a₁,b₁; a₂,b₂; ...) with n people has n equations x_i + a_i = b_i(T   x_i - a_i), which can be rewritten as x_i + a_i(1+b_i) = b_i(T - x_i), so we see that this is the same problem as discussed in 7.R.1 below where men find a purse, but with variable known purses, p_i = a_i(1+b_i). We get x_i = b_iT/(1+b_i) - a_i. Adding these for all i gives one equation in the one unknown T, T [Σ {b_i/(1+b_i} - 1] = Σ a_i.

For II (a₁,b₁; a₂,b₂; ...), systematic elimination in the n equations x_i + a_i = b_i (x_i+1   a_i) leads to x₁ [b₁b₂...b_n   1]   =  a₁(b₁+1) + a₂b₁(b₂+1) + a₃b₁b₂(b₃+1) + .... , and any other value can be found by shifting the starting point of the cycle.

Verse versions: Euclid; Wingate/Kersey; Ozanam; Vinot;

Euclid. c-325. Opera. Ed. by J. L. Heiberg & H. Menge, Teubner, Leipzig, 1916. Vol. VIII, pp. 286 287. Ass and mule in Greek and Latin verse. (1, 2; 1, 1). (Sanford 207 gives English of Clavius's 1605 version. Cf Wingate/Kersey.)

Diophantos. Arithmetica. c250. Book I.