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

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(We can safely divide by <math>n</math> because it must be positive). So the total number of players <math>p</math> can only be <math>3</math>, <math>6</math>, and <math>9</math>.  
 
(We can safely divide by <math>n</math> because it must be positive). So the total number of players <math>p</math> can only be <math>3</math>, <math>6</math>, and <math>9</math>.  
  
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Since the total number of games <math>p(p-1)/2  is </math>(1.4+1 = 12/5)<math> times a non-negative integer number of games won by righties, </math>p(p-1)<math> must be a multiple of </math>12(2) = 24<math>. Among \{</math>3,6,9<math>\}, only </math>p = 9<math> satisfies this condition,  so the total number of games is </math>(9)(8)/2 = \boxed{\textbf{(B)}\ 36}.$
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Since the total number of games <math>p(p-1)/2</math> is <math>(1.4+1 = 12/5)</math> times a non-negative integer number of games won by righties, <math>p(p-1)</math> must be a multiple of <math>12(2) = 24</math>. Among <math>\{3,6,9\}</math>, only <math>p = 9</math> satisfies this condition,  so the total number of games is <math>(9)(8)/2 = \boxed{\textbf{(B)}\ 36}.</math>
  
 
~ggao5uiuc, oinava, yingkai_0_ (minor edits)
 
~ggao5uiuc, oinava, yingkai_0_ (minor edits)

Revision as of 20:29, 6 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 answer. 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 $p$ can only be $3$, $6$, and $9$.

Since the total number of games $p(p-1)/2$ is $(1.4+1 = 12/5)$ times a non-negative integer number of games won by righties, $p(p-1)$ must be a multiple of $12(2) = 24$. Among $\{3,6,9\}$, only $p = 9$ satisfies this condition, so the total number of games is $(9)(8)/2 = \boxed{\textbf{(B)}\ 36}.$

~ggao5uiuc, oinava, 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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