2023 AMC 12A Problems/Problem 13

Problem

In a table tennis tournament every participant played every other participant exactly once. Although there were twice as many right-handed players as left-handed players, the number of games won by left-handed players was $40\%$ more than the number of games won by right-handed players. (There were no ties and no ambidextrous players.) What is the total number of games played?

$\textbf{(A) }15\qquad\textbf{(B) }36\qquad\textbf{(C) }45\qquad\textbf{(D) }48\qquad\textbf{(E) }66$

Solution 1 (3 min solve)

We know that the total amount of games must be the sum of games won by left and right handed players. Then, we can write $g = l + r$ , and since $l = 1.4r$ , $g = 2.4r$ . Given that $r$ and $g$ are both integers, $g/2.4$ also must be an integer. From here we can see that $g$ must be divisible by 12, leaving only answers B and D. Now we know the formula for how many games are played in this tournament is $n(n-1)/2$ , the sum of the first $n-1$ triangular numbers. Now setting 36 and 48 equal to the equation will show that two consecutive numbers must equal 72 or 96. Clearly $72=8*9$ , so the answer is $\boxed{\textbf{(B) }36}$ .

~~ Antifreeze5420

Solution 2

First, every player played the other, so there's $n\choose2$ games. Also, if the right-handed won $x$ games, the left handed won $7/5x$ , meaning that the total amount of games was $12/5x$ , so the total amount of games is divisible by $12$ .

Then, we do something funny and look at the answer choices. Only $\boxed{\textbf{(B) }36}$ satisfies our 2 findings.

Solution 3

Let $r$ be the amount of games the right-handed won. Since the left-handed won $1.4r$ games, the total number of games played can be expressed as $(2.4)r$ , or $12/5r$ , meaning that the answer is divisible by 12. This brings us down to two answer choices, $B$ and $D$ . We note that the answer is some number $x$ choose $2$ . This means the answer is in the form $x(x-1)/2$ . Since answer choice D gives $48 = x(x-1)/2$ , and $96 = x(x-1)$ has no integer solutions, we know that $\boxed{\textbf{(B) }36}$ is the only possible choice.

Video Solution 1 by OmegaLearn

https://youtu.be/BXgQIV2WbOA

2023 AMC 10A (Problems • Answer Key • Resources)
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2023 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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All AMC 12 Problems and Solutions

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

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