Dudeney, Amusements in Mathematics Problem 17 Four brothers - named John, William, Charles and Thomas - had each a money-box. They boxes were all given to them on the same day, and they had at once put what money they had into them; only, as the boxes were not very large, they first changed the money into as few coins as possible. After they had done this, they told one another how much money they had saved, and it was found that if John had had 2s more in his box than at present, if William had had 2s less, if Charles had had twice as much, and if Thomas had had half as much, they would all have had exactly the same amount. Now, when I add that all four boxes together contained 45s, and that there were only six coins in all in them, it becomes an entertaining puzzle to discover just what coins were in each box.
Let the amounts be J, W, C, T for John, William, Charles, Thomas respectively. We are given that J+2 = W-2 = 2C = Â½T. So J = 2C-2, W = 2C+2, T = 4C. So J+W+C+T = 9C = 45, so C = 5, J = 8, W = 12, T = 20. We can obviously use 1 coin each for C (crown) and T (sovereign). W can be done with 2 coins (half- sovereign + florin). That leaves two coins for J, which can be done, provided we allow double-florins. Dudeney 1-25 (C) John Scholes jscholes@kalva.demon.co.uk 25 Oct 2002