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

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.

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;

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.

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;

Diophantos. Arithmetica. c250. Book I.