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Difference between revisions of "2023 AMC 12A Problems/Problem 13"
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Let <math>r</math> be the amount of games the right-handed won. Since the left-handed won <math>1.4r</math> games, the total number of games played can be expressed as <math>(2.4)r</math>, or <math>12/5r</math>, meaning that the answer is divisible by 12. This brings us down to two answer choices, <math>B</math> and <math>D</math>.  
 
Let <math>r</math> be the amount of games the right-handed won. Since the left-handed won <math>1.4r</math> games, the total number of games played can be expressed as <math>(2.4)r</math>, or <math>12/5r</math>, meaning that the answer is divisible by 12. This brings us down to two answer choices, <math>B</math> and <math>D</math>.  
 
We note that the answer is some number <math>x</math> choose <math>2</math>. This means the answer is in the form <math>x(x-1)/2</math>. Since answer choice D gives <math>48 = x(x-1)/2</math>, and <math>96 = x(x-1)</math> has no integer solutions, we know that <math>\boxed{\textbf{(B) }36}</math> is the only possible choice.
 
We note that the answer is some number <math>x</math> choose <math>2</math>. This means the answer is in the form <math>x(x-1)/2</math>. Since answer choice D gives <math>48 = x(x-1)/2</math>, and <math>96 = x(x-1)</math> has no integer solutions, we know that <math>\boxed{\textbf{(B) }36}</math> is the only possible choice.
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==Solution 3==
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Here is the rigid way to prove that 36 is the only result. Let the number of left-handed players is n, so the right-handed palyer is 2n. The number of games won by the left-handed players comes in two ways: (1) the games played by two left-left paris, which is <math>\frac{n(n-1)}{2}</math>, and (2) the games played by left-right pairs, which is x. And <math>x\leq 2n^2</math>. so <cmath>\frac{\frac{n(n-1)}{2}+x}{\frac{2n(2n-1)}{2}+2n^2-x}=1.4</cmath>, which gives <cmath>x=\frac{17n^2}{8}-\frac{3n}{8}</cmath>, by setting up the inequality <math>x\leq 2n^2</math>, it comes <math>n\leq 3</math>. So the total number of player can only be <math>3</math>, <math>6</math>, and <math>9</math>. Then we can rule out all other possible values for total number of games they played.
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== Video Solution 1 by OmegaLearn ==
 
== Video Solution 1 by OmegaLearn ==

Revision as of 20:34, 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.

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 have a product of 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. Therefore, that rules out D. 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 $12$. 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 $\boxed{\textbf{(B) }36}$

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.

Solution 3

Here is the rigid way to prove that 36 is the only result. Let the number of left-handed players is n, so the right-handed palyer is 2n. The number of games won by the left-handed players comes in two ways: (1) the games played by two left-left paris, which is $\frac{n(n-1)}{2}$, and (2) the games played by left-right pairs, which is x. And $x\leq 2n^2$. so \[\frac{\frac{n(n-1)}{2}+x}{\frac{2n(2n-1)}{2}+2n^2-x}=1.4\], which gives \[x=\frac{17n^2}{8}-\frac{3n}{8}\], by setting up the inequality $x\leq 2n^2$, it comes $n\leq 3$. So the total number of player can only be $3$, $6$, and $9$. Then we can rule out all other possible values for total number of games they played.


Video Solution 1 by OmegaLearn

https://youtu.be/BXgQIV2WbOA

Video Solution 2 by TheBeautyofMath

https://www.youtube.com/watch?v=sLtsF1k9Fx8&t=227s

Video Solution

https://youtu.be/S9l1C6pjXWI

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

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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