Difference between revisions of "2021 AIME I Problems/Problem 7"
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If <math>k = 1</math>, then <math>m \equiv n \equiv 1 \pmod 4</math>. This corresponds to choosing two elements from the set <math>\{1, 5, 9, 13, 17, 21, 25, 29\}</math>. There are <math>\binom 82</math> ways to do so. | If <math>k = 1</math>, then <math>m \equiv n \equiv 1 \pmod 4</math>. This corresponds to choosing two elements from the set <math>\{1, 5, 9, 13, 17, 21, 25, 29\}</math>. There are <math>\binom 82</math> ways to do so. | ||
| − | If <math>k < 1</math>, by multiplying <math>m</math> and <math>n</math> by the same constant <math>c = \frac{1}{k}</math>, we have that <math>mc \equiv nc \equiv 1 \pmod 4</math>. | + | If <math>k < 1</math>, by multiplying <math>m</math> and <math>n</math> by the same constant <math>c = \frac{1}{k}</math>, we have that <math>mc \equiv nc \equiv 1 \pmod 4</math>. Then either <math>m \equiv n \equiv 1 \pmod 4</math>, or <math>m \equiv n \equiv 3 \pmod 4</math>. But the first case was already counted, so we don't need to consider that case. The other case corresponds to choosing two numbers from the set <math>\{3, 7, 11, 15, 19, 23, 27\}</math>. There are <math>\binom 72</math> ways here. |
Finally, if <math>k > 1</math>, note that <math>k</math> must be an integer. This means that <math>m, n</math> belong to the set <math>\{k, 5k, 9k, \dots\}</math>, or <math>\{3k, 7k, 11k, \dots\}</math> (why?). Taking casework on <math>k</math>, we get the sets <math>\{2, 10, 18, 26\}, \{6, 14, 22, 30\}, \{4, 20\}, \{12, 28\}</math>. Some sets have been omitted; this is because they were counted in the other cases already. This sums to <math>\binom 42 + \binom 42 + \binom 22 + \binom 22</math>. | Finally, if <math>k > 1</math>, note that <math>k</math> must be an integer. This means that <math>m, n</math> belong to the set <math>\{k, 5k, 9k, \dots\}</math>, or <math>\{3k, 7k, 11k, \dots\}</math> (why?). Taking casework on <math>k</math>, we get the sets <math>\{2, 10, 18, 26\}, \{6, 14, 22, 30\}, \{4, 20\}, \{12, 28\}</math>. Some sets have been omitted; this is because they were counted in the other cases already. This sums to <math>\binom 42 + \binom 42 + \binom 22 + \binom 22</math>. | ||
Revision as of 20:53, 11 March 2021
Problem
Find the number of pairs 
of positive integers with 
such that there exists a real number 
satisfying![\[\sin(mx)+\sin(nx)=2.\]](http://latex.artofproblemsolving.com/3/d/c/3dc7189bf3226f56ef411df7b580c6778ace2c72.png)
Solution
The maximum value of 
is 
, which is achieved at 
for some integer 
. This is left as an exercise to the reader.
This implies that 
, and that 
and 
, for integers 
.
Taking their ratio, we have ![\[\frac{mx}{nx} = \frac{\frac{\pi}{2}+2a\pi}{\frac{\pi}{2}+2b\pi} \implies \frac{m}{n} = \frac{4a + 1}{4b + 1} \implies \frac{m}{4a + 1} = \frac{n}{4b + 1} = k.\]](http://latex.artofproblemsolving.com/1/e/c/1ec583b832f97ee0791a483adae7cfc39f83cda2.png)
It remains to find all 
that satisfy this equation.
If 
, then 
. This corresponds to choosing two elements from the set 
. There are 
ways to do so.
If 
, by multiplying 
and 
by the same constant 
, we have that 
. Then either , or 
. But the first case was already counted, so we don't need to consider that case. The other case corresponds to choosing two numbers from the set 
. There are 
ways here.
Finally, if 
, note that must be an integer. This means that belong to the set 
, or 
(why?). Taking casework on , we get the sets 
. Some sets have been omitted; this is because they were counted in the other cases already. This sums to 
.
In total, there are 
pairs of 
.
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. 
