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Doubling the others

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$4$ players A, B, C and D play $4$ rounds. In round $1$ player A pays each of the other $3$ players as many coins as that player then holds, doubling each of them. In round $2$ player B does the same, in round $3$ player C, and in round $4$ player D. After round $4$ every player holds $32$ coins. How many coins did player A hold at the start?

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Hint

Undo the rounds from the last: the payer's opponents held half as much before that round.

Solution

Undoing round $4$ : A, B and C had $16$ each, and D had $32 + 3 \times 16 = 80$ .
Undoing round $3$ : A and B had $8$ each, D had $40$ , and C had $16 + 8 + 8 + 40 = 72$ .
Undoing round $2$ : A had $4$ , C had $36$ , D had $20$ , and B had $8 + 4 + 36 + 20 = 68$ .
Undoing round $1$ : B had $34$ , C had $18$ , D had $10$ , and A had $4 + 34 + 18 + 10 = 66$ .
So player A started with $66$ coins.

Answer: $66$

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