AIME 2013 I, Problem 8

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: 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 $5368 + 3 = 5371 \equiv \fbox{371} \pmod{1000}$.

Answer: 371