2021 AIME I Problems/Problem 7

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 1

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 $mc \equiv nc \equiv 1 \pmod 4$ . Then either $m \equiv n \equiv 1 \pmod 4$ , or $m \equiv n \equiv 3 \pmod 4$ . 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 $\{3, 7, 11, 15, 19, 23, 27\}$ . 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 $\{3k, 7k, 11k, \dots\}$ . 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~

Solution 2

In order for $\sin(mx) + \sin(nx) = 2$ , $\sin(mx) = \sin(nx) = 1$ .

This happens when $mx \equiv nx \equiv \frac{\pi}{2} ($ mod $2\pi).$

This means that $mx = \frac{\pi}{2} + 2\pi\alpha$ and $nx = \frac{\pi}{2} + 2\pi\beta$ for any integers $\alpha$ and $\beta$ .

As in Solution 1, take the ratio of the two equations: $\frac{mx}{nx} = \frac{\frac{\pi}{2}+2\pi\alpha}{\frac{\pi}{2}+2\pi\beta} \implies \frac{m}{n} = \frac{\frac{1}{2}+2\alpha}{\frac{1}{2}+2\beta} \implies \frac{m}{n} = \frac{4\alpha+1}{4\beta+1}$

Now notice that the numerator and denominator of $\frac{4\alpha+1}{4\beta+1}$ are both odd, which means that $m$ and $n$ have the same power of two (the powers of 2 cancel out).

Let the common power be $p$ : then $m = 2^p\cdot a$ , and $n = 2^p\cdot b$ where $a$ and $b$ are integers between 1 and 30.

We can now rewrite the equation: $\frac{2^p\cdot a}{2^p\cdot b} = \frac{4\alpha+1}{4\beta+1} \implies \frac{a}{b} = \frac{4\alpha+1}{4\beta+1}$

Now it is easy to tell that $a \equiv 1 ($ mod $4)$ and $b \equiv 1 ($ mod $4)$ . However, there is another case: that

$a \equiv 3 ($ mod $4)$ and $b \equiv 3 ($ mod $4)$ . This is because multiplying both $a$ and $b$ by $-1$ will not change the fraction, but each congruence will be changed to $-1 ($ mod $4) \equiv 3 ($ mod $4)$ .

From the first set of congruences, we find that $a$ and $b$ can be two of $\{1, 5, 9, \cdots, 29\}$ .

From the second set of congruences, we find that $a$ and $b$ can be two of $\{3, 7, 11, \cdots, 27\}$ .

Now all we have to do is multiply by $2^p$ to get back to $m$ and $n$ . Let’s organize the solutions in order of increasing values of $p$ , keeping in mind that $m$ and $n$ are bounded between 1 and 30.

For $p = 0$ we get $\{1, 5, 9, \cdots, 29\}, \{3, 7, 11, \cdots, 27\}$ .

For $p = 1$ we get $\{2, 10, 18, 26\}, \{6, 14, 22, 30\}$

For $p = 2$ we get $\{4, 20\}, \{12, 28\}$

If we increase the value of $p$ more, there will be less than two integers in our sets, so we are done there.

There are 8 numbers in the first set, 7 in the second, 4 in the third, 4 in the fourth, 2 in the fifth, and 2 in the sixth.
In each of these sets we can choose 2 numbers to be  and  and then assign them in increasing order. Thus there are:

$\dbinom{8}{2}+\dbinom{7}{2}+\dbinom{4}{2}+\dbinom{4}{2}+\dbinom{2}{2}+\dbinom{2}{2}$

$= 28+21+6+6+1+1 = \box63$

\begin{align*} &p & \text{Possible Solutions For $m, n$ } \\ &0 & $\{1, 5, 9, \cdots, 29\}, \{3, 7, 11, \cdots, 27\}$ \\ &1 & $\{2, 10, 18, 26\}, \{6, 14, 22, 30\}$ \\ &2 & $\{4, 20\}, \{12, 28\}$ \\ \end{align*}

I WILL FINISH THE SOLUTION SOON, PLEASE DO NOT EDIT THIS BEFORE THEN, THANK YOU!

-KingRavi

2021 AIME I (Problems • Answer Key • Resources)
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2021 AIME I Problems/Problem 7

Contents

Problem

Solution 1

Solution 2

See also