# 2013 AIME I Problems/Problem 8

## Problem

The domain of the function $f(x) = \arcsin(\log_{m}(nx))$ is a closed interval of length $\frac{1}{2013}$ , where $m$ and $n$ are positive integers and $m>1$. Find the remainder when the smallest possible sum $m+n$ is divided by 1000.

## Solution 1

We know that the domain of $\text{arcsin}$ is $[-1, 1]$, so $-1 \le \log_m nx \le 1$. Now we can apply the definition of logarithms: $$m^{-1} = \frac1m \le nx \le m$$ $$\implies \frac{1}{mn} \le x \le \frac{m}{n}$$ Since the domain of $f(x)$ has length $\frac{1}{2013}$, we have that $$\frac{m}{n} - \frac{1}{mn} = \frac{1}{2013}$$ $$\implies \frac{m^2 - 1}{mn} = \frac{1}{2013}$$

A larger value of $m$ will also result in a larger value of $n$ since $\frac{m^2 - 1}{mn} \approx \frac{m^2}{mn}=\frac{m}{n}$ meaning $m$ and $n$ increase about linearly for large $m$ and $n$. So we want to find the smallest value of $m$ that also results in an integer value of $n$. The problem states that $m > 1$. Thus, first we try $m = 2$: $$\frac{3}{2n} = \frac{1}{2013} \implies 2n = 3 \cdot 2013 \implies n \notin \mathbb{Z}$$ Now, we try $m=3$: $$\frac{8}{3n} = \frac{1}{2013} \implies 3n = 8 \cdot 2013 \implies n = 8 \cdot 671 = 5368$$ Since $m=3$ is the smallest value of $m$ that results in an integral $n$ value, we have minimized $m+n$, which is $5368 + 3 = 5371 \equiv \boxed{371} \pmod{1000}$.

## Solution 2

We start with the same method as above. The domain of the arcsin function is $[-1, 1]$, so $-1 \le \log_{m}(nx) \le 1$. $$\frac{1}{m} \le nx \le m$$ $$\frac{1}{mn} \le x \le \frac{m}{n}$$ $$\frac{m}{n} - \frac{1}{mn} = \frac{1}{2013}$$ $$n = 2013m - \frac{2013}{m}$$

For $n$ to be an integer, $m$ must divide $2013$, and $m > 1$. To minimize $n$, $m$ should be as small as possible because increasing $m$ will decrease $\frac{2013}{m}$, the amount you are subtracting, and increase $2013m$, the amount you are adding; this also leads to a small $n$ which clearly minimizes $m+n$.

We let $m$ equal $3$, the smallest factor of $2013$ that isn't $1$. Then we have $n = 2013*3 - \frac{2013}{3} = 6039 - 671 = 5368$ $m + n = 5371$, so the answer is $\boxed{371}$.

Note that we need $-1\le f(x)\le 1$, and this eventually gets to $\frac{m^2-1}{mn}=\frac{1}{2013}$. From there, break out the quadratic formula and note that $$m= \frac{n+\sqrt{n^2+4026^2}}{2013\times 2}.$$ Then we realize that the square root, call it $a$, must be an integer. Then $(a-n)(a+n)=4026^2.$
Observe carefully that $4026^2 = 2\times 2\times 3\times 3\times 11\times 11\times 61\times 61$! It is not difficult to see that to minimize the sum, we want to minimize $n$ as much as possible. Seeing that $2a$ is even, we note that a $2$ belongs in each factor. Now, since we want to minimize $a$ to minimize $n$, we want to distribute the factors so that their ratio is as small as possible (sum is thus minimum). The smallest allocation of $2, 61, 61$ and $2, 11, 3, 3, 11$ fails; the next best is $2, 61, 11, 3, 3$ and $2, 61, 11$, in which $a=6710$ and $n=5368$. That is our best solution, upon which we see that $m=3$, thus $\boxed{371}$.
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 