Difference between revisions of "2004 AMC 12A Problems/Problem 7"
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==Solution== | ==Solution== | ||
− | + | 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. | |
− | + | 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 , , and start with , , and tokens, respectively. How many rounds will there be in the game?
Solution
We look at a set of three rounds, where the players begin with , , and tokens. After three rounds, there will be a net loss of token per player (they receive two tokens and lose three). Therefore, after rounds -- or three-round sets, and will have , , and tokens, respectively. After more round, player will give away tokens, leaving it empty-handed, and thus the game will end. We then have there are 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.