A QUEER COINCIDENCE.

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 thirty-two cents
In his pocket. The puzzle is to find out how
Much money each man had with him before he sat down to play.

Let A0, B0,. G0 be how much money each had before the start. For Adam: He won the first game, so he was giving away money after the first game. During the next 6 games he was doubling his money every game. x*(2^6)=32, here x - the money Adam was left with after the first game.

The first equation: A0-B0-C0-D0-E0-F0-G0=0.5 cents.

The second equation: 2*B0-0.5-2*(C0+D0+E0+F0+G0)=1 cent (The B doubled his money after the first game, gave away some money after the second game, and kept doubling the next 5 games, hence he was left with 1 cent after his win.)

If you keep building equations like this, you will end up with a system of 7 linear equations of 7 uknowns, which should have a straightforward solution. HTH