# Difference between revisions of "2012 AMC 10B Problems/Problem 15"

## 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

The total number of games in the tournament is $\frac{6 \times 5}{2}= 15$. Here's a 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 its 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

Look above. Since there are $15$ wins, and $15$ losses in each of these games, you look and see that $\frac{15}{3}=5$, so $5$ teams can each have 3 wins, and since all the wins have been accounted for, you know that the last team must have lost all of their games.

## See Also

 2012 AMC 10B (Problems • Answer Key • Resources) Preceded byProblem 14 Followed byProblem 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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