Dudeney, Amusements in Mathematics Problem 24 Two ladies went into a shop, where, through some curious eccentricity, no change was given, and made purchases amounting together to less than five shillings. "Do you know," said one lady, "I find I shall require no fewer than six current coins of the realm to pay for what I have bought." The other lady considered a moment, and then exclaimed: "By a peculiar coincidence, I am exactly in the same dilemma." "Then we will pay the two bills together." But, to their astonishment, they still required six coins. What is the smallest possible amount of their purchases - both different? Solution
: 11Â¾d and 1s 11Â½d. Let us first consider what amounts require 6 coins. Obviously we should have an exact amount of pence + Â¾d to require a Â½d and a Â¼d. It is easy to check that all round amounts from 1d to 11d can be managed with 1 or 2 coins except 5d, 10d which both require 3 coins and 11d which requires 4. [We assume the 2d and 4d coins were no longer current.] So the smallest amounts requiring 6 coins are 11Â¾d, 1s 5Â¾d, 1s 10Â¾d, 1s 11Â¼ and 1s 11Â½d. We have 11Â¾d + 1s 5Â¾d = 2s 5Â½d, which only requires 5 coins (2s + 3d + 1d + 1d + Â½d). Similarly, 11Â¾d + 1s 10Â¾d = 2s 10Â½d, which only requires 5 coins (2s + 6d + 3d + 1d + Â½d) and 11Â¾d + 1s 11Â¼d = 2s 11d, which only requires 5 coins (2s + 6d + 3d + 1d + 1d). But 11Â¾d + 1s 11Â½d = 2s 11Â¼d, which requires 6 coins (2s + 6d + 3d + 1d + 1d + Â¼d). Similarly, 1s 5Â¾d + 1s 10Â¾d = 3s 4Â½d, which requires (2s + 1s + 3d + 1d + Â½d). Curiously, Dudeney's answer is wrong (1s 5Â¾d and 1s 11Â½d). Dudeney 1-25 (C) John Scholes jscholes@kalva.demon.co.uk 25 Oct 2002