2004 AMC 8 Problems/Problem 5: Difference between revisions

Revision as of 23:55, 4 July 2013

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

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. $\boxed{\textbf{(D)}\ 15}$ .

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

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 ==See Also==
 {{AMC8 box|year=2004|num-b=4|num-a=6}}
+{{MAA Notice}}

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