8. The "Reverse" Game. Seven friends decided to play cards by somewhat unusual rules. The one who won had to pay each of the other players as much money as they had in their pocket. The players played seven rounds and, strangely enough, they won in alphabetical order of their names, starting with $A, B, C, D, E, F$ and $G$.

After finishing the game, the friends discovered that each of them had exactly 1 dollar and 28 cents left. How much money did each player have before the game started?
8. Players \(A, B, C, D, E, F\) and \(G\) had, before the game began, 4 dollars 49 cents, 2 dollars 25 cents, 1 dollar 13 cents, 57 cents, 29 cents, 15 cents, and 8 cents, respectively. The answer can be obtained by working from the end of the problem to the beginning, but a simpler method is as follows: \(7+1=8 ; 2 \times 7+1=15 ; 4 \times 7+1=29\) and so on (the first multipliers represent successive powers of two, that is, the numbers \(2,4,8,16,32\) and 64\()\).