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Difference between revisions of "2015 AMC 12A Problems/Problem 13"

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==Solution==
 
==Solution==
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We can eliminate answer choices <math>\textbf{(A)}</math> and <math>\textbf{(B)}</math> because there are an even number of scores, so if one is false, the other must be false too.
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We can eliminate answer choices <math>\textbf{(A)}</math> and <math>\textbf{(B)}</math> because there are an even number of scores, so if one is false, the other must be false too. Answer choice <math>\textbf{(C)}</math> must be true since every team plays every other team, so it is impossible for two teams to lose every game. Answer choice <math>\textbf{(D)}</math> must be true since each game gives out a total of two points, and there are <math>\binom{12}{2} = 66</math> games, for a total of <math>132</math> points. Answer choice <math>\boxed{\textbf{(E)}}</math> is false. If everyone draws each of their 11 games, then every team will tie for first place with 11 points each.
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Answer choice <math>\textbf{(C)}</math> must be true since every team plays every other team, so it is impossible for two teams to lose every game.
 
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Answer choice <math>\textbf{(D)}</math> must be true since each game gives out a total of two points, and there are <math>\frac{11\times 12}{2} = 66</math> games, for a total of <math>132</math> points.
 
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Answer choice <math>\boxed{(E)}</math> is false (and thus our answer). If everyone draws each of their 11 games, then every team will tie for first place with 11 points each.
 
  
 
== See Also ==
 
== See Also ==
 
{{AMC12 box|year=2015|ab=A|num-b=12|num-a=14}}
 
{{AMC12 box|year=2015|ab=A|num-b=12|num-a=14}}

Revision as of 17:04, 15 December 2018

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.

See Also

2015 AMC 12A (Problems • Answer Key • Resources)
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
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