Problem 36

Since $m$ and $n$ are positive integers with $n > 1$ and $m^{n} = 2^{25} \times 3^{40}$, then 2 and 3 are prime factors of $m$ (since they are prime factors of $m^{n}$) and must be the only prime factors of $m$ (since if there were other prime factors of $m$, then there would be other prime factors of $m^{n}$). Therefore, $m = 2^{a} \times 3^{b}$ for some positive integers $a$ and $b$ and so $m^{n} = (2^{a} \times 3^{b})^{n} = 2^{an} \times 3^{bn}$. Since $m^{n} = 2^{25} \times 3^{40}$, then we must have $an = 25$ and $bn = 40$. Since $a, b, n$ are positive integers, then $n$ is a common divisor of 25 and 40. Since $n > 1$, then $n = 5$, which means that $a = 5$ and $b = 8$. In this case, $m = 2^{5} \times 3^{8} = 32 \times 6561 = 209952$, which gives $m + n = 209952 + 5 = 209957$.