Art of Problem Solving

2004 AMC 8 Problems/Problem 5: Difference between revisions

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<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 1==
===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.
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===
==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>.
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>.



Revision as of 17:52, 26 October 2016

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?

$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 16$

Solution 1

The remaining team will be the only undefeated one. The other $\boxed{\textbf{(D)}\ 15}$ teams must have lost a game before getting out, thus fifteen games yielding fifteen losers.

Solution 2

There will be $8$ games the first round, $4$ games the second round, $2$ games the third round, and $1$ game in the final round, giving us a total of $8+4+2+1=15$ games. $\boxed{\textbf{(D)}\ 15}$.

See Also

2004 AMC 8 (ProblemsAnswer KeyResources)
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

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