### Instruction: 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 the smallest possible value of $m+n.$.### Response: I notice that the function $f(x)$ is composed of three functions: $\arcsin(x)$, $\log_{m}(x)$, and $nx$.
I know that the domain of $\arcsin(x)$ is $[-1, 1]$, which means that $\log_{m}(nx)$ must also be in that interval.
I also know that the domain of $\log_{m}(x)$ is $(0, \infty)$, which means that $nx$ must be positive.
Therefore, I can conclude that the domain of $f(x)$ is the intersection of the intervals $(-\frac{1}{n}, \frac{1}{n})$ and $(0, \infty)$, which is $(0, \frac{1}{n})$.
The length of this interval is $\frac{1}{n}$, which is given to be $\frac{1}{2013}$.
This implies that $n = 2013$.
Now, I need to find the smallest possible value of $m$.
I recall that the base of a logarithm must be greater than 1, so the smallest possible value of $m$ is 2.
Therefore, the smallest possible value of $m+n$ is $2 + 2013 = 2015$.