2023 AMC 12A Problems/Problem 13
- The following problem is from both the 2023 AMC 10A #16 and 2023 AMC 12A #13, so both problems redirect to this page.
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2
- 4 Solution 3
- 5 Solution 4
- 6 Solution 5 (🧀Cheese🧀)
- 7 Video Solution by Power Solve
- 8 Video Solution
- 9 Video Solution 1 by OmegaLearn
- 10 Video Solution by CosineMethod
- 11 Video Solution 2 by TheBeautyofMath
- 12 Video Solution
- 13 Video Solution by Math-X
- 14 Video Solution by SpreadTheMathLove
- 15 See Also
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
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 have a product of 72 or 96. Clearly, , so the answer is .
~~ Antifreeze5420
Solution 2
First, we know that every player played every other player, so there's a total of games since each pair of players forms a bijection to a game. Therefore, that rules out D. Also, if we assume the right-handed players won a total of games, the left-handed players must have won a total of games, meaning that the total number of games played was . Thus, the total number of games must be divisible by . Therefore leaving only answer choices B and D. Since answer choice D doesn't satisfy the first condition, the only answer that satisfies both conditions is
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.
Solution 4
Here is the rigid way to prove that 36 is the only result. Let the number of left-handed players be n, so the right-handed player is 2n. The number of games won by the left-handed players comes in two ways: (1) the games played by two left-left pairs, which is , and (2) the games played by left-right pairs, which is x. And . so which gives By setting up the inequality , it comes . So the total number of players can only be , , and . Then we can rule out all other possible values for the total number of games they played.
~ ~ ggao5uiuc
~ ~ yingkai_0_ (minor edits)
Solution 5 (🧀Cheese🧀)
If there are players, the total number of games played must be , so it has to be a triangular number. The ratio of games won by left-handed to right-handed players is , so the number of games played must also be divisible by . Finally, we notice that only satisfies both of these conditions.
~MathFun1000
Video Solution by Power Solve
Video Solution
https://youtu.be/wSeSYxDCOZ8 ~Education, the Study of Everything
Video Solution 1 by OmegaLearn
Video Solution by CosineMethod
https://www.youtube.com/watch?v=N-eZMOv_ZjY
Video Solution 2 by TheBeautyofMath
https://www.youtube.com/watch?v=sLtsF1k9Fx8&t=227s
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution by Math-X
https://youtu.be/N2lyYRMuZuk?si=7IEmYDGBHi5i_XKt&t=1452 ~Math-X
Video Solution by SpreadTheMathLove
https://www.youtube.com/watch?v=sypOvNiR3sw
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.