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| − | ==Problem==
| + | #redirect[[2023 AMC 12A Problems/Problem 13]] |
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| − | 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 <math>40\%</math> 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?
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| − | <math>\textbf{(A) }15\qquad\textbf{(B) }36\qquad\textbf{(C) }45\qquad\textbf{(D) }48\qquad\textbf{(E) }66</math>
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| − | ==Solution 1 (3 min solve)==
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| − | 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 <math>g = l + r</math>, and since <math>l = 1.4r</math>, <math>g = 2.4r</math>. Given that <math>r</math> and <math>g</math> are both integers, <math>g/2.4</math> also must be an integer. From here we can see that <math>g</math> 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 <math>n(n-1)/2</math>, the sum of the first <math>n-1</math> triangular numbers. Now setting 36 and 48 equal to the equation will show that two consecutive numbers must equal 72 or 96. Clearly <math>72=8*9</math>, so the answer is <math>\boxed{(B)}</math>.
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| − | ~~ Antifreeze5420
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| − | ==Solution 2==
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| − | First, every player played the other, so there's <math>n\choose2</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>.
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| − | Then, we do something funny and look at the answer choices. Only <math>\boxed{B}</math> satisfies our 2 findings.
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| − | == Video Solution 1 by OmegaLearn ==
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| − | https://youtu.be/BXgQIV2WbOA
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| − | ==See Also==
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| − | {{AMC10 box|year=2023|ab=A|num-b=15|num-a=17}}
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| − | {{MAA Notice}}
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