A QUEER COINCIDENCE.
(
Money Puzzles)
Seven men, whose names were Adams, Baker, Carter, Dobson, Edwards,
Francis, and Gudgeon, were recently engaged in play. The name of the
particular game is of no consequence. They had agreed that whenever a
player won a game he should double the money of each of the other
players--that is, he was to give the players just as much money as they
had already in their pockets. They played seven games, and, strange to
say, each won a game in turn, in the order in which their names are
given. But a more curious coincidence is this--that when they had
finished play each of the seven men had exactly the same amount--two
shillings and eightpence--in his pocket. The puzzle is to find out how
much money each man had with him before he sat down to play.
Answer:
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 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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