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 "2021 AIME I Problems/Problem 7"
m
Line 18: Line 18:
 
In total, there are <math>\binom 82 + \binom 72 + \binom 42 + \binom 42 + \binom 22 + \binom 22 = \boxed{63}</math> pairs of <math>(m, n)</math>.
 
In total, there are <math>\binom 82 + \binom 72 + \binom 42 + \binom 42 + \binom 22 + \binom 22 = \boxed{63}</math> pairs of <math>(m, n)</math>.
  
−
This solution was brought to you by Leonard_my_dude.
+
This solution was brought to you by ~Leonard_my_dude~.
 
==See also==
 
==See also==
 
{{AIME box|year=2021|n=I|num-b=6|num-a=8}}
 
{{AIME box|year=2021|n=I|num-b=6|num-a=8}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 20:41, 11 March 2021

Problem

Find the number of pairs $(m,n)$ of positive integers with $1\le m<n\le 30$ such that there exists a real number $x$ satisfying\[\sin(mx)+\sin(nx)=2.\]

Solution

The maximum value of $\sin \theta$ is $1$, which is achieved at $\theta = \frac{\pi}{2}+2k\pi$ for some integer $k$. This is left as an exercise to the reader.

This implies that $\sin(mx) = \sin(nx) = 1$, and that $mx = \frac{\pi}{2}+2a\pi$ and $nx = \frac{\pi}{2}+2b\pi$, for integers $a, b$.

Taking their ratio, we have \[\frac{mx}{nx} = \frac{\frac{\pi}{2}+2a\pi}{\frac{\pi}{2}+2b\pi} \implies \frac{m}{n} = \frac{4a + 1}{4b + 1} \implies \frac{m}{4a + 1} = \frac{n}{4b + 1} = k.\] It remains to find all $m, n$ that satisfy this equation.

If $k = 1$, then $m \equiv n \equiv 1 \pmod 4$. This corresponds to choosing two elements from the set $\{1, 5, 9, 13, 17, 21, 25, 29\}$. There are $\binom 82$ ways to do so.

If $k < 1$, by multiplying $m$ and $n$ by the same constant $c = \frac{1}{k}$, we have that $mc \equiv nc \equiv 1 \pmod 4$. This means $m \equiv n \equiv 3 \pmod 4$. This corresponds to choosing two numbers from the set $\{3, 7, 11, 15, 19, 23, 27\}$. There are $\binom 72$ ways here.

Finally, if $k > 1$, note that $k$ must be an integer. This means that $m, n$ belong to the set $\{k, 5k, 9k, \dots\}$, or $\{3k, 7k, 11k, \dots\}$ (why?). Taking casework on $k$, we get the sets $\{2, 10, 18, 26\}, \{6, 14, 22, 30\}, \{4, 20\}, \{12, 28\}$. Some sets have been omitted; this is because they were counted in the other cases already. This sums to $\binom 42 + \binom 42 + \binom 22 + \binom 22$.

In total, there are $\binom 82 + \binom 72 + \binom 42 + \binom 42 + \binom 22 + \binom 22 = \boxed{63}$ pairs of $(m, n)$.

This solution was brought to you by ~Leonard_my_dude~.

See also

2021 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME 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=2021_AIME_I_Problems/Problem_7&oldid=149130"