Difference between revisions of "2023 AMC 12A Problems/Problem 13"

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==Solution 2==
 
==Solution 2==
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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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. 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>.  
  
Then, we do something funny and look at the answer choices. Only <math>\boxed{\textbf{(B) }36}</math> satisfies our 2 findings.
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Then, we examine the answer choices and see which choices are divisible by <math>12</math>. It turns out that only <math>\boxed{\textbf{(B) }36}</math> satisfies our constraints.
  
 
==Solution 3==
 
==Solution 3==

Revision as of 00:32, 10 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 $\boxed{\textbf{(B) }36}$.

~~ Antifreeze5420

Solution 2

First, we know that every player played every other player, so there's a total of $\dbinom{n}{2}$ games, since each pair of players forms a bijection to a game. Also, if we assume the right-handed players won a total of $x$ games, the left-handed players must have won a total of $\dfrac{7}{5}x$ games, meaning that the total number of games played was $\dfrac{12}{5}x$. Thus, the total number of games must be divisible by $12$.

Then, we examine the answer choices and see which choices are divisible by $12$. It turns out that only $\boxed{\textbf{(B) }36}$ satisfies our constraints.

Solution 3

Let $r$ be the amount of games the right-handed won. Since the left-handed won $1.4r$ games, the total number of games played can be expressed as $(2.4)r$, or $12/5r$, meaning that the answer is divisible by 12. This brings us down to two answer choices, $B$ and $D$. We note that the answer is some number $x$ choose $2$. This means the answer is in the form $x(x-1)/2$. Since answer choice D gives $48 = x(x-1)/2$, and $96 = x(x-1)$ has no integer solutions, we know that $\boxed{\textbf{(B) }36}$ is the only possible choice.

Video Solution 1 by OmegaLearn

https://youtu.be/BXgQIV2WbOA

See Also

2023 AMC 10A (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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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