Difference between revisions of "2004 AMC 12A Problems/Problem 7"
m ({{duplicate}}) |
(→See also) |
||
Line 13: | Line 13: | ||
{{AMC12 box|year=2004|ab=A|num-b=6|num-a=8}} | {{AMC12 box|year=2004|ab=A|num-b=6|num-a=8}} | ||
{{AMC10 box|year=2004|ab=A|num-b=7|num-a=9}} | {{AMC10 box|year=2004|ab=A|num-b=7|num-a=9}} | ||
+ | {{MAA Notice}} |
Revision as of 19:15, 3 July 2013
- 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 3 rounds, where the players have , , and 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, repectively. After 1 more round, player will give away his last 3 tokens and the game will stop .
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.