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Weird Game Puzzle

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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.
strange to say, each won a game in turn in the order of their names
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?

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JamesBond Pundit Asked on 28th July 2015 in Microsoft Interview Puzzles.
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    • 1 Answer(s)

    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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    ravi Expert Answered on 28th July 2015.
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