Difference between revisions of "2004 AMC 8 Problems/Problem 5"

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== Problem ==
 
== 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?
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Ms. Hamilton's eighth-grade class wants to participate in the annual three-person-team basketball tournament. 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 ==
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==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===
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==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>.
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2004|num-b=4|num-a=6}}
 
{{AMC8 box|year=2004|num-b=4|num-a=6}}
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{{MAA Notice}}

Latest revision as of 22:03, 6 March 2021

Problem

Ms. Hamilton's eighth-grade class wants to participate in the annual three-person-team basketball tournament. 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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