2004 AMC 12A Problems/Problem 7

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Problem

A game is played with tokens according to the following rule. In each round, the player with the most tokens gives one token to each of the other players and also places one token in the discard pile. The game ends when some player runs out of tokens. Players $A$, $B$, and $C$ start with 15, 14, and 13 tokens, respectively. How many rounds will there be in the game?

$\mathrm{(A) \ } 36 \qquad \mathrm{(B) \ } 37 \qquad \mathrm{(C) \ } 38 \qquad \mathrm{(D) \ } 39 \qquad \mathrm{(E) \ } 40$

Solution

Look at a set of 3 rounds, where the players have $x+1$, $x$, and $x-1$ tokens. Each of the players will gain two tokens from the others and give away 3 tokens, so overall, each player will lose 1 token.

Therefore, after 12 sets of 3 rounds, or 36 rounds, the players will have 3, 2, and 1 tokens, repectively. After 1 more round, player $A$ will give away his last 3 tokens and the game will stop $\Rightarrow\mathrm{(B)}$.

See also

2004 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions
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