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.


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 1

We look at a set of three rounds, where the players begin with $x+1$, $x$, and $x-1$ tokens. After three rounds, there will be a net loss of $1$ token per player (they receive two tokens and lose three). Therefore, after $36$ rounds -- or $12$ three-round sets, $A,B$ and $C$ will have $3$, $2$, and $1$ tokens, respectively. After $1$ more round, player $A$ will give away $3$ tokens, leaving them empty-handed, and thus the game will end. We then have there are $36+1=\boxed{\mathrm{(B)}\ 37}$ rounds until the game ends.

Solution 2

Let's bash a few rounds. The amounts are for players $1,2,$ and $3$, respectively.

First round: $15,14,13$ (given) Second round: $12,15,14$ Third round: $13,12,15$ Fourth round: $14,13,12$

We see that after $3$ rounds are played, we have the exact same scenario as the first round but with one token less per player. So, the sequence $1,4,7,10...$ where each of the next members are $3$ greater than the previous one corresponds with the sequence $15,14,13,12...$ where the first sequence represents the round and the second sequence represents the number of tokens player $1$ has. But we note that once player $1$ reaches $3$ coins, the game will end on his next turn as he must give away all his coins. Therefore, we want the $15-3+1=13$th number in the sequence $1,4,7,10...$ which is $\boxed{\mathrm{(B)}\ 37}$.

Solution by franzliszt

Solution 3

Looking at a set of five rounds, you'll see $A$ has $4$ fewer tokens than in the beginning. Looking at four more rounds, you'll notice $A$ has the same amount of tokens, namely $11$, compared to round five. If you keep doing this process, you'll see a pattern: Every four rounds, the amount of tokens $A$ has either decreased by $4$ or stayed the same compared to the previous four rounds. For example, in round nine, $A$ had $11$ tokens, in round $13$, $A$ had $11$ tokens, and in round $17$, $A$ had $7$ tokens, etc. Using this weird pattern, you can find out that in round $37$, $A$ should have $3$ tokens, but since they would have given them away in that round, the game would end on $\boxed{\mathrm{(B)}\ 37}$.

This strategy is a very slow one, requiring you to list out about $25$ rounds.

Video Solution


Education, the Study of Everything

See also

2004 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
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 12 Problems and Solutions
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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