Dudeney, Amusements in Mathematics Problem 25 The common cash consists of brass coins of varying thickness, with a round, square, or triangular hold in the center. These are strung on wires like buttons. Supposing that eleven coins with round holes are worth fifteen ching-changs, that eleven with square holes are worth sixteen ching-changs, and that eleven with triangular holes are worth seventeen ching-changs, how can a Chinaman give me change for half a crown, using no coins other than the three mentioned? A ching-chang is worth exactly twopence and four-fifteenths of a ching-chang.
Let the ching-chang have value C. Then C = 2d + 4C/15, so C = 30d/11. Thus we require coins with a total value of 11C. Let the value of the round, square, triangular holed coins be R, S, T respectively. Then 11R = 15C, 11S = 16C, 11T = 17C. Suppose we use x Rs, y Ss and z Ts. Then 15x/11 + 16y/11 + 17z/11 = 11. So 15x + 16y + 17z = 121. It is not hard to check that the only solution is x = 7, y = 1, z = 0. Dudeney 1-25 (C) John Scholes jscholes@kalva.demon.co.uk 25 Oct 2002