If $m$ and $n$ are positive integers with $n>1$ such that $m^{n}=2^{25} \times 3^{40}$, then $m+n$ is
(A) 209962
(B) 1954
(C) 209957
(D) 6598
(E) 1049760

## Part C: Each correct answer is worth 8.
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 $m^{n}=\left(2^{a} \times 3^{b}\right)^{n}=2^{a n} \times 3^{b n}$. Since $m^{n}=2^{25} \times 3^{40}$, then we must have $a n=25$ and $b n=40$.

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 $m+n=209952+5=209957$.