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

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==Solution==
 
==Solution==
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Look at a set of <math>3</math> rounds, where the players have <math>x+1</math>, <math>x</math>, and <math>x-1</math> tokens. Each of the players will gain two tokens from the others and give away <math>3</math> tokens, so overall, each player will lose <math>1</math> token. Therefore, after <math>12</math> sets of <math>3</math> rounds, or <math>36</math> rounds, the players will have <math>3</math>, <math>2</math>, and <math>1</math> tokens, respectively.  After <math>1</math> more round, player <math>A</math> will give away his last <math>3</math> tokens and the game will end. <math>\boxed{\mathrm{(B)}\ 37}</math>  
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We look at a set of three rounds, where the players begin with <math>x+1</math>, <math>x</math>, and <math>x-1</math> tokens.
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(I hope this makes sense I can't explain it any better)
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After three rounds, there will be a net loss of <math>1</math> token per player (they receive two tokens and lose three). Therefore, after <math>36</math> rounds -- or <math>12</math> three-round sets, <math>A,B</math> and <math>C</math> will have <math>3</math>, <math>2</math>, and <math>1</math> tokens, respectively.  After <math>1</math> more round, player <math>A</math> will give away <math>3</math> tokens, leaving it empty-handed, and thus the game will end. We then have there are <math>36+1=\boxed{\mathrm{(B)}\ 37}</math> rounds until the game ends.
  
 
== See also ==
 
== See also ==

Revision as of 17:29, 4 January 2019

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

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 it empty-handed, and thus the game will end. We then have there are $36+1=\boxed{\mathrm{(B)}\ 37}$ rounds until the game ends.

See also

2004 AMC 12A (Problems • Answer Key • Resources)
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 (Problems • Answer Key • Resources)
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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