Difference between revisions of "2004 AMC 12A Problems/Problem 7"
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==Solution== | ==Solution== | ||
| − | Look at a set of 3 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 3 tokens, so overall, each player will lose 1 token. | + | 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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| − | Therefore, after 12 sets of 3 rounds, or 36 rounds, the players will have 3, 2, and 1 tokens, | ||
== See also == | == See also == | ||
Revision as of 23:24, 20 July 2014
- 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 
, 
, and 
start with 15, 14, and 13 tokens, respectively. How many rounds will there be in the game?
Solution
Look at a set of 
rounds, where the players have 
, 
, and 
tokens. Each of the players will gain two tokens from the others and give away tokens, so overall, each player will lose 
token. Therefore, after 
sets of rounds, or 
rounds, the players will have , 
, and tokens, respectively. After more round, player will give away his last tokens and the game will end. 
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
