# Difference between revisions of "2004 AMC 12A Problems/Problem 7"

The following problem is from both the 2004 AMC 12A #7 and 2004 AMC 10A #8, so both problems redirect to this page.

## 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, respectively. After $1$ more round, player $A$ will give away his last $3$ tokens and the game will end. $\boxed{\mathrm{(B)}\ 37}$ (I hope this makes sense I can't explain it any better)