Difference between revisions of "2023 AMC 10A Problems/Problem 16"

Revision as of 20:38, 9 November 2023

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 $\boxed{(B)}$ .

~~ Antifreeze5420

Solution 2

First, every player played the other, so there's $n\c2$ 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 $\boxed{B}$ satisfies our 2 findings.

Video Solution 1 by OmegaLearn

https://youtu.be/BXgQIV2WbOA

2023 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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All AMC 10 Problems and Solutions

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

@@ Line 9: / Line 9: @@
 ~~ Antifreeze5420
+==Solution 2==
+First, every player played the other, so there's <math>n\c2</math> games. Also, if the right-handed won <math>x</math> games, the left handed won <math>7/5x</math>, meaning that the total amount of games was <math>12/5x</math>, so the total amount of games is divisible by <math>12</math>.
+Then, we do something funny and look at the answer choices. Only <math>\boxed{B}</math> satisfies our 2 findings.
 == Video Solution 1 by OmegaLearn ==

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