(related to Problem: A Queer Coincidence)

Puzzles of this class are generally solved in the old books by the tedious process of "working backwards." But a simple general solution is as follows: If there are $n$ players, the amount held by every player at the end will be $m(2^n),$ the last winner must have held $m(n+1)$ at the start, the next $m(2n+1),$ the next $m(4n+1),$ the next $m(8n+1),$ and so on to the first player, who must have held $m(2^{n-1}n+1).$

Thus, in this case, n = 7, and the amount held by every player at the end was $27$ farthings. Therefore $m = 1,$ and $G$ started with $8$ farthings, $F$ with $15,$ $E$ with $29,$ $D$ with $57,$ $C$ with $113,$ $B$ with $225,$ and $A$ with $449$ farthings.

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Project Gutenberg

  1. Dudeney, H. E.: "Amusements in Mathematics", The Authors' Club, 1917

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