2023 AMC 12A Problems/Problem 13
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[hide]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 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?
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 , and since , . Given that and are both integers, also must be an integer. From here we can see that 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 , the sum of the first triangular numbers. Now setting 36 and 48 equal to the equation will show that two consecutive numbers must equal 72 or 96. Clearly , so the answer is .
~~ Antifreeze5420
Solution 2
First, every player played the other, so there's games. Also, if the right-handed won games, the left handed won , meaning that the total amount of games was , so the total amount of games is divisible by .
Then, we do something funny and look at the answer choices. Only satisfies our 2 findings.
Solution 3
Let be the amount of games the right-handed won. Since the left-handed won games, the total number of games played can be expressed as , or , meaning that the answer is divisible by 12. This brings us down to two answer choices, and . We note that the answer is some number choose . This means the answer is in the form . Since answer choice D gives , and has no integer solutions, we know that is the only possible choice.
Video Solution 1 by OmegaLearn
See Also
2023 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
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 10 Problems and Solutions |
2023 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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