Difference between revisions of "2012 AMC 10B Problems/Problem 15"
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<math> \textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ 6 </math> | <math> \textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ 6 </math> | ||
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The total number of games in the tournament is <math>\frac{6 \times 5}{2}= 15</math>. | The total number of games in the tournament is <math>\frac{6 \times 5}{2}= 15</math>. | ||
Here's a poorly done chart of 15 games: | Here's a poorly done chart of 15 games: | ||
Revision as of 19:51, 7 December 2015
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?
Solution
The total number of games in the tournament is 
.
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 
See Also
| 2012 AMC 10B (Problems • Answer Key • Resources) | ||
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
