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Difference between revisions of "2021 AIME I Problems/Problem 7"

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In total, there are <math>\boxed{62}</math> possible solutions.
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In total, there are <math>\boxed{062}</math> possible solutions.
  
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However the answer is <math>63</math>, where is the last possible solution?
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However the answer is <math>063</math>, where is the last possible solution?
  
 
~Interstigation
 
~Interstigation

Revision as of 19:02, 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

Since $-1\leq\sin(x)\leq1$, $\sin(mx)+\sin(nx)=2$ means that each of $\sin(mx)$ and $\sin(nx)$ must be exactly $1$. Then $m$ and $n$ must be cycles away, or the difference between them must be multiple of $4$. If $m$ is $1$, then $n$ can be $5,9,13,17,21,25,29$. Like this, the table below can be listed:

Range of $m$ Number of Possible $n$'s
Case 1 $1 \leq m \leq 2$ $7$
Case 2 $3 \leq m \leq 6$ $6$
Case 3 $7 \leq m \leq 10$ $5$
Case 4 $11 \leq m \leq 14$ $4$
Case 5 $15 \leq m \leq 18$ $3$
Case 6 $19 \leq m \leq 22$ $2$
Case 7 $23 \leq m \leq 26$ $1$
Case 8 $27 \leq m \leq 30$ $0$

In total, there are $\boxed{062}$ possible solutions.

However the answer is $063$, where is the last possible solution?

~Interstigation

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

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