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

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==Problem==
 
==Problem==
 
Find the number of pairs <math>(m,n)</math> of positive integers with <math>1\le m<n\le 30</math> such that there exists a real number <math>x</math> satisfying<cmath>\sin(mx)+\sin(nx)=2.</cmath>
 
Find the number of pairs <math>(m,n)</math> of positive integers with <math>1\le m<n\le 30</math> such that there exists a real number <math>x</math> satisfying<cmath>\sin(mx)+\sin(nx)=2.</cmath>
 +
 +
==Solution==
 +
Since <math>-1\leq\sin(x)\leq1</math>, <math>\sin(mx)+\sin(nx)=2</math> means that each of <math>\sin(mx)</math> and <math>\sin(nx)</math> must be exactly <math>1</math>. Then <math>m</math> and <math>n</math> must be cycles away, or the difference between them must be multiple of <math>4</math>. If <math>m</math> is <math>1</math>, then <math>n</math> can be <math>5,9,13,17,21,25,29</math>. Like this, the table below can be listed:
 +
 +
{| class="wikitable" style="text-align:center;width:100%"
 +
|-
 +
|
 +
! scope="col" | '''Range of <math>m</math>'''
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! scope="col" | '''Number of Possibilities'''
 +
|-
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! scope="row" | '''Case 1'''
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| <math>1 \leq m \leq 2</math>
 +
| <math>7</math>
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|-
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! scope="row" | '''Case 2'''
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| <math>3 \leq m \leq 6</math>
 +
| <math>6</math>
 +
|-
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! scope="row" | '''Case 3'''
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| <math>7 \leq m \leq 10</math>
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| <math>5</math>
 +
|-
 +
! scope="row" | '''Case 4'''
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| <math>11 \leq m \leq 14</math>
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| <math>4</math>
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|-
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! scope="row" | '''Case 5'''
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| <math>15 \leq m \leq 18</math>
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| <math>3</math>
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|-
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! scope="row" | '''Case 6'''
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| <math>19 \leq m \leq 22</math>
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| <math>2</math>
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|-
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! scope="row" | '''Case 7'''
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| <math>23 \leq m \leq 26</math>
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| <math>1</math>
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|-
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! scope="row" | '''Case 8'''
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| <math>27 \leq m \leq 30</math>
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| <math>0</math>
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|-
 +
|}
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 +
In total, there are <math>\boxed{62}</math> possible solutions.
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 +
However the answer is <math>63</math>, where is the last possible solution?
 +
 +
~Interstigation
  
 
==Solution==
 
==Solution==

Revision as of 19:57, 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 Possibilities
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 $\boxed{62}$ possible solutions.

However the answer is $63$, where is the last possible solution?

~Interstigation

Solution

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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