Prob. 11, 12, 14, pp. 2 & 59; 1895?: 12, 13, 15, pp. 8 & 63; 1917: 13, 12, 17, pp. 8 & 57. (1, 2; 1, 1). (3, 1; 3, 2). (5, 2; 5, 1)

No. 21, pp. 156 & 333. (20, 4; 20, 3/2).

No. 11, pp. 163 & 334. II-(20, 4; 40, 4/9; 60, 7/4). Answer:  (5620, 1880, 6700)/19

Hummerston. Fun, Mirth & Mystery. 1924. Pocket money, Puzzle no. 4, pp. 21 & 172. (0, 2; ½, 3).

Sullivan. Unusual. 1947. Prob. 36: A problem old enough to be considered new. (1, 1; 1, 2).

David Singmaster. Some diophantine recreations. Op. cit. in 7.P.5. 1993. Sketches some history; finds condition for integer data in (a, b; c, d) to produce an integer solution, namely (bd - 1)/(b+1, d+1) divides a + c where (b+1, d+1) is the Greatest Common Divisor of b+1 and d+1. A letter from S. Parameswaran interpreted the problem given in Alcuin as though the second statement was also made by the first animal. This gives x + a = b (y - a); x + c = d (y - c) and similar reasoning finds the integrality conditions for this variant.

David Singmaster. A variation of the ass and mule problem. CM 28: 4 (May 2002) 236-238. The 2001 Maritime Mathematics Contest had a specific case of the following. The first person says: "If I had a from you, I'd have b times you, but if I gave c to you, I'd have d times you." This leads to the equations: x + a = b (y - a); d (y + c) = x - c. The integral d of the classic problem has been changed to 1/d with d integral. Note that b > d for reasonable solutions. Reasoning similar to the previous article finds the condition for integer data to produce an integer solution and the condition is simpler than for that problem.

Tomislav Došlić. Fibonacci in Hogwarts? MG 87 (No. 510) (Nov 2003) 432-436. Observes that Fibonacci does the case (7, 5; 5, 7), whose solution is non-integral: (121, 167)/17. He considers the case (a, b; b, a) and finds there are 16 positive pairs a, b which give integral solutions. If we set a  b, these are a, b = 1, 2; 1, 3; 1, 5; 2, 2; 2, 3; 2, 8; 3, 3; 3, 7; 5, 8.

David Singmaster. Integral solutions of ass and mule problems. Gives a simpler solution for Došlić's result and finds all integral solutions when positivity is not required.
7.R.1. MEN FIND A PURSE AND 'BLOOM OF THYMARIDAS'
See Tropfke 604 & 606.

Algebraically, 7.R.1 and 7.R.2 differ only in signs.

NOTATION. Finding a purse has two forms.

I (a₁, a₂, ..., a_n) -- i-th says "If I had the purse, I'd have a_i times the rest of you".

II (a₁, a₂, ..., a_n) -- i-th says "If I had the purse, I'd have a_i times the i+1-st person".

There are two related problems which I call forms III and IV.

III (a₁, a₂, ..., a_n) -- the sum of all amounts except the i th is a_i. See the discussion under

Iamblichus. A number of problems in 7.H lead to this type of problem when one uses

reciprocal times as work rates.

IV-(a₁, a₂, ..., a_n) -- x_i + x_i+1 = a_i. (This is determinate only if n is odd.)

For n = 3, types III and IV are the same, though the constants or the variables are taken in a different order, so that III (a, b, c) = IV-(c, a, b) if we keep the variables in the same order.

I give answers as a list of the amounts, in order; then the purse.

Let p be the value of the purse and let T be the total of the amounts. Then I (a₁,a₂,...) with n people has n equations x_i + p = a_i(T   x_i), so we see that this is the same problem as discussed in 7.R.2 below where men buy a horse, but with the value of the horse and the multipliers all negative, which makes this version have fewer sign complications in its solution. Thus we get x_i = (a_iT-p)/(1+a_i). Adding these for all i gives one equation in the two unknowns T and p. However, letting C = T + p leads to the simplest equation: (n 1)T  =  [Σ  1/(1+a_i)] C.

For II (a₁,a₂,...), systematic elimination in the n equations x_i + p = a_ix_i+1 leads to x₁ [a₁a₂...a_n - 1] = p [1 + a₁ + a₁a₂ + ... + a₁a₂...a_n], and any other value can be found by shifting the starting point of the cycle.

In either case, the solution can be adapted to variable purses -- see 7.R. In some problems, gaining the purse is replaced by paying out, so the purse can be treated as having a negative value -- see Unger, 1838.

Other versions of the problem has several horses and a saddle (or other equipage) or several cups and a cover. The i-th horse with the saddle is worth a_i times the others or the next. I don't seem to have entered any of these until recently.

There are versions where one asks for a fraction x/y such that (x+a)/y = b, x/(y+c) = d, where a, c are integral (possibly non-positive) and b, d are rationals. This is a mixture between 7.R and 7.R.1, but fits better here as we can think of a, c as purses. I will denote this as:

V-(a, b; c, d). This has solution x = β(a + αb)/(α - β); y = (a + βb)/(α - β).

See: Dodson; Todhunter.

Some of the cistern problems in 7.H are of type III

No. 16, p. 135. "To find three numbers such that the sums of pairs are given numbers." He does III (20, 30, 40).

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