Difference between revisions of "2023 AMC 12A Problems/Problem 13"
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==Solution 2== | ==Solution 2== | ||
| − | First, we know that every player played every other player, so there's a total of <math>\dbinom{n}{2}</math> 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 <math>x</math> games, the left-handed players must have won a total of <math>\dfrac{7}{5}x</math> games, meaning that the total number of games played was <math>\dfrac{12}{5}x</math>. Thus, the total number of games must be divisible by <math>12</math>. 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 | + | First, we know that every player played every other player, so there's a total of <math>\dbinom{n}{2}</math> 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 <math>x</math> games, the left-handed players must have won a total of <math>\dfrac{7}{5}x</math> games, meaning that the total number of games played was <math>\dfrac{12}{5}x</math>. Thus, the total number of games must be divisible by <math>12</math>. 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 <math>\boxed{\textbf{(B) }36}</math> |
==Solution 3== | ==Solution 3== | ||
Revision as of 17:50, 10 November 2023
- The following problem is from both the 2023 AMC 10A #16 and 2023 AMC 12A #13, so both problems redirect to this page.
Contents
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, 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.
Video Solution 1 by OmegaLearn
Video Solution 2 by TheBeautyofMath
https://www.youtube.com/watch?v=sLtsF1k9Fx8&t=227s
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. 
