**A
QUEER COINCIDENCE.— solution**

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*(2*n*+1), the next
*m*(4*n*+1), the next *m*(8*n*+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 2^{7} 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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