Difference between revisions of "2004 AMC 8 Problems/Problem 5"
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The losing team of each game is eliminated from the tournament. If sixteen teams compete, how many games will be played to determine the winner? | The losing team of each game is eliminated from the tournament. If sixteen teams compete, how many games will be played to determine the winner? | ||
| − | <math> \textbf{(A)}4\qquad\textbf{(B)}7\qquad\textbf{(C)}8\qquad\textbf{(D)}15\qquad\textbf{(E)}16 </math> | + | <math> \textbf{(A)}\ 4\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 16 </math> |
== Solution == | == Solution == | ||
| − | There will be <math>8</math> games the first round, <math>4</math> games the second round, <math>2</math> games the third round, and <math>1</math> game in the final round, giving us a total of <math>8+4+2+1=15</math> games. <math>\boxed{\textbf{(D)}\ 15}</math> | + | ===Solution 1=== |
| + | The remaining team will be the only undefeated one. The other <math>\boxed{\textbf{(D)}\ 15}</math> teams must have lost a game before getting out, thus fifteen games yielding fifteen losers. | ||
| + | |||
| + | ===Solution 2=== | ||
| + | There will be <math>8</math> games the first round, <math>4</math> games the second round, <math>2</math> games the third round, and <math>1</math> game in the final round, giving us a total of <math>8+4+2+1=15</math> games. <math>\boxed{\textbf{(D)}\ 15}</math>. | ||
| + | |||
| + | ==See Also== | ||
| + | {{AMC8 box|year=2004|num-b=4|num-a=6}} | ||
Revision as of 04:32, 24 December 2012
Problem
The losing team of each game is eliminated from the tournament. If sixteen teams compete, how many games will be played to determine the winner?
Solution
Solution 1
The remaining team will be the only undefeated one. The other 
teams must have lost a game before getting out, thus fifteen games yielding fifteen losers.
Solution 2
There will be 
games the first round, 
games the second round, 
games the third round, and 
game in the final round, giving us a total of 
games. .
See Also
| 2004 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 4 |
Followed by Problem 6 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
