2012 AMC 10B Problems/Problem 15

Revision as of 17:24, 28 December 2014 by Lee541 (talk | contribs) (Solution 1)

Problem

In a round-robin tournament with 6 teams, each team plays one game against each other team, and each game results in one team winning and one team losing. At the end of the tournament, the teams are ranked by the number of games won. What is the maximum number of teams that could be tied for the most wins at the end on the tournament?

$\textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ 6$

Solution 1

The total number of games in the tournament is 6*5/2= 15. Here's a poorly done chart of 15 games:

|  1  2  3  4  5  6 |
|1 X  W  L  W  L  W |
|2 L  X  W  L W  W | 
|3 W  L  X  W  L  W |
|4 L  W  L  X  W  W |
|5 W  L  W  L  X  W |
|6 L  L  L  L  L  X |

The "X's" are for when it is where a team is set against itself, which cannot happen. The chart says that Team 6 has lost all of it's matches, which means that each of the other teams won against it. Then, alternating Wins and Losses were tried. It shows that it is possible for 5 teams to tie for the same amount of wins, which in this case is 3 wins. Thus, the answer is $\boxed{\textbf{(D)}\ 5}$

Solution 2

The total number of games in the tournament is 6*5/2= 15. Now, we see which numbers from 1-6 divide 15. 1, 3, and 5 divide 15. 5 is the largest of the 3 numbers, and it is possible for each of 5 teams to win 3 games. Thus, the answer is $\boxed{\textbf{(D)}\ 5}$

See Also

2012 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
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 AMC 10 Problems and Solutions

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