Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2023 AMC 12A Problems/Problem 13"
(→‎Solution 3)
Line 22: Line 22:
 
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.
 
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.
  
 +
 +
[[User:Ggao5uiuc|Ggao5uiuc]] ([[User talk:Ggao5uiuc|talk]])ggao5uiuc
  
 
== Video Solution 1 by OmegaLearn ==
 
== Video Solution 1 by OmegaLearn ==

Revision as of 20:35, 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.


Ggao5uiuc (talk)ggao5uiuc

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2023_AMC_12A_Problems/Problem_13&oldid=202114"