Question: Seven men engaged in play. Whenever a player won a game he doubled the money of each of the other players. That is, he gave each player just as much money as each had in his pocket. They played seven games and, strange to say, each won a game in turn in the order of their names, which
began with the letters A, B, C, D, E, F, and G.
When they had finished it was found that each man had exactly $1.28 in his pocket. How much had each man in his pocket before play?
Answer: The seven men, A, B, C, D, E, F, and G, had respectively in their pockets before play the following sums: $4.49, $2.25, $1.13, 57¢, 29¢, 15¢, and 8¢. The answer may be found by laboriously working backwards, but a simpler method is as follows: 7 + 1 = 8; 2 X 7 + 1 = 15; 4 X 7 + 1 = 29; and so on, where the multiplier increases in powers of 2, that is, 2, 4, 8, 16, 32, and 64.
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