Art of Problem Solving

2004 AMC 8 Problems/Problem 5: Difference between revisions

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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?
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 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.
Note that the winning team will the be the only team that wins all of the games. Therefore, to find the total number of games to determine the winner has a 1:1 correspondence to the number of ways to determine the losers. ( Think of it this way: If you want to select two balls from a bag of 6 balls, it is analogous to selecting the 4 balls that you don't want to select, both are 6 choose 2.). There are 15 losing teams, and since each round is unique, there are 15 total rounds.  
 
~Brackie1331


==Solution 2==
==Solution 2==

Latest revision as of 09:37, 16 May 2024

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

Note that the winning team will the be the only team that wins all of the games. Therefore, to find the total number of games to determine the winner has a 1:1 correspondence to the number of ways to determine the losers. ( Think of it this way: If you want to select two balls from a bag of 6 balls, it is analogous to selecting the 4 balls that you don't want to select, both are 6 choose 2.). There are 15 losing teams, and since each round is unique, there are 15 total rounds.

~Brackie1331

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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