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Difference between revisions of "2023 AMC 12A Problems/Problem 13"
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==Solution 4==
 
==Solution 4==
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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 <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 players can only be <math>3</math>, <math>6</math>, and <math>9</math>. Then we can rule out all other possible values for the total number of games they played.
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Here is the rigid way to prove that <math>36</math> is the only result. Let the number of left-handed players be <math>n</math>, so the number of right-handed players is <math>2n</math>. The number of games won by the left-handed players comes in two ways:  
 +
*The games played by two left-left pairs, which is <math>\frac{n(n-1)}{2}</math>, and  
 +
*The games played by left-right pairs, which we'll call <math>x</math>.  
 +
Note that <math>x\leq 2n^2,</math> which is the total number of games played by left-right pairs. Using the same logic for right-right pairs and right-left pairs, we have that <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>  
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We know that <math>x\leq 2n^2</math>, applying that becomes
 +
<cmath>\begin{align*}
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\frac{17n^2}{8}-\frac{3n}{8} &\le 2n^2 \\
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17n^2 - 3n &\le 16n^2 \\
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n^2 - 3n &\le 0 \\
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n &\le 3.
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\end{align*}</cmath>
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(We can safely divide by <math>n</math> because it must be positive). So the total number of players can only be <math>3</math>, <math>6</math>, and <math>9</math>. Using this information we can rule out all answer choices except <math>\boxed{\textbf{(B)}\ 36}.</math>
  
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~ggao5uiuc, yingkai_0_ (minor edits)
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~ ~ ggao5uiuc
 
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~ ~ yingkai_0_ (minor edits)
 
  
 
==Solution 5 (🧀Cheese🧀)==
 
==Solution 5 (🧀Cheese🧀)==
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~MathFun1000
 
~MathFun1000
  
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==Video Solution (⚡Under 2 min⚡)==
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== Video Solution by Power Solve ==
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https://youtu.be/2_ytwADrIto
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 +
==Video Solution ==
 
https://youtu.be/wSeSYxDCOZ8
 
https://youtu.be/wSeSYxDCOZ8
 
<i> ~Education, the Study of Everything </i>
 
<i> ~Education, the Study of Everything </i>
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== Video Solution 0 by Power Solve (easy to understand!) ==
 
−
https://youtu.be/2_ytwADrIto
 
  
 
== Video Solution 1 by OmegaLearn ==
 
== Video Solution 1 by OmegaLearn ==
 
https://youtu.be/BXgQIV2WbOA
 
https://youtu.be/BXgQIV2WbOA
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 +
== Video Solution by CosineMethod ==
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https://www.youtube.com/watch?v=N-eZMOv_ZjY
  
 
== Video Solution 2 by TheBeautyofMath ==
 
== Video Solution 2 by TheBeautyofMath ==
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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
  
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==Video Solution by Math-X (First understand the problem!!!)==
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==Video Solution by Math-X ==
 
https://youtu.be/N2lyYRMuZuk?si=7IEmYDGBHi5i_XKt&t=1452
 
https://youtu.be/N2lyYRMuZuk?si=7IEmYDGBHi5i_XKt&t=1452
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~Math-X
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~Math-X
  
 
==Video Solution by SpreadTheMathLove==
 
==Video Solution by SpreadTheMathLove==

Revision as of 15:49, 5 August 2024

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

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 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 number of right-handed players is $2n$. The number of games won by the left-handed players comes in two ways:

  • The games played by two left-left pairs, which is $\frac{n(n-1)}{2}$, and
  • The games played by left-right pairs, which we'll call $x$.

Note that $x\leq 2n^2,$ which is the total number of games played by left-right pairs. Using the same logic for right-right pairs and right-left pairs, we have that \[\frac{\frac{n(n-1)}{2}+x}{\frac{2n(2n-1)}{2}+2n^2-x}=1.4,\] which gives \[x=\frac{17n^2}{8}-\frac{3n}{8}.\] We know that $x\leq 2n^2$, applying that becomes \begin{align*} \frac{17n^2}{8}-\frac{3n}{8} &\le 2n^2 \\ 17n^2 - 3n &\le 16n^2 \\ n^2 - 3n &\le 0 \\ n &\le 3. \end{align*} (We can safely divide by $n$ because it must be positive). So the total number of players can only be $3$, $6$, and $9$. Using this information we can rule out all answer choices except $\boxed{\textbf{(B)}\ 36}.$

~ggao5uiuc, yingkai_0_ (minor edits)

Solution 5 (🧀Cheese🧀)

If there are $n$ players, the total number of games played must be $\binom{n}{2}$, so it has to be a triangular number. The ratio of games won by left-handed to right-handed players is $7:5$, so the number of games played must also be divisible by $12$. Finally, we notice that only $\boxed{\textbf{(B) }36}$ satisfies both of these conditions.

~MathFun1000

Video Solution by Power Solve

https://youtu.be/2_ytwADrIto

Video Solution

https://youtu.be/wSeSYxDCOZ8 ~Education, the Study of Everything

Video Solution 1 by OmegaLearn

https://youtu.be/BXgQIV2WbOA

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

https://youtu.be/S9l1C6pjXWI

~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

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