2015 AMC 12A Problems/Problem 13

Revision as of 23:42, 3 October 2021 by Whatisit (talk | contribs) (Solution)

Problem

A league with 12 teams holds a round-robin tournament, with each team playing every other team exactly once. Games either end with one team victorious or else end in a draw. A team scores 2 points for every game it wins and 1 point for every game it draws. Which of the following is NOT a true statement about the list of 12 scores?

$\textbf{(A)}\ \text{There must be an even number of odd scores.}\\ \qquad\textbf{(B)}\ \text{There must be an even number of even scores.}\\ \qquad\textbf{(C)}\ \text{There cannot be two scores of }0\text{.}\\ \qquad\textbf{(D)}\ \text{The sum of the scores must be at least }100\text{.}\\ \qquad\textbf{(E)}\ \text{The highest score must be at least }12\text{.}$

Solution

We can eliminate answer choices $\textbf{(A)}$ and $\textbf{(B)}$ because there are an even number of scores, so if one is false, the other must be false too. Answer choice $\textbf{(C)}$ must be true since every team plays every other team, so it is impossible for two teams to lose every game. Answer choice $\textbf{(D)}$ must be true since each game gives out a total of two points, and there are $\binom{12}{2} = 66$ games, for a total of $132$ points. Answer choice $\boxed{\textbf{(E)}}$ is false. If everyone draws each of their 11 games, then every team will tie for first place with 11 points each. Quick question, why does the answer key say 3?

See Also

2015 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
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 12 Problems and Solutions