Two positive integers $m$ and $n$ are chosen such that $m$ is the smallest positive integer with only two positive divisors and $n$ is the largest integer less than $100$ with exactly three positive divisors. What is $m+n$?
I know that the smallest positive integer with only two positive divisors is 2, because any smaller number would have either zero or one divisor. I also know that the only way an integer can have exactly three positive divisors is if it is a perfect square of a prime number, because any other factorization would give more than three divisors. So I need to find the largest prime number less than $10$, since its square will be less than $100$. The largest prime number less than $10$ is $7$, so the largest integer less than $100$ with exactly three positive divisors is $7^2=49$. Therefore, $m=2$ and $n=49$, and $m+n=51$. 