How many [positive integers](https://artofproblemsolving.com/wiki/index.php/Positive_integer) have exactly three [proper divisors](https://artofproblemsolving.com/wiki/index.php/Proper_divisor) (positive integral [divisors](https://artofproblemsolving.com/wiki/index.php/Divisor) excluding itself), each of which is less than 50?
Suppose $n$ is such an [integer](https://artofproblemsolving.com/wiki/index.php/Integer). Because $n$ has $3$ proper divisors, it must have $4$ divisors,, so $n$ must be in the form $n=p\cdot q$ or $n=p^3$ for distinct [prime numbers](https://artofproblemsolving.com/wiki/index.php/Prime_number) $p$ and $q$. 
In the first case, the three proper divisors of $n$ are $1$, $p$ and $q$.  Thus, we need to pick two prime numbers less than $50$. There are fifteen of these ($2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43$ and $47$) so there are ${15 \choose 2} =105$ ways to choose a pair of primes from the list and thus $105$ numbers of the first type.
In the second case, the three proper divisors of $n$ are 1, $p$ and $p^2$.  Thus we need to pick a prime number whose square is less than $50$.  There are four of these ($2, 3, 5,$ and $7$) and so four numbers of the second type. 
Thus there are $105+4=\boxed{109}$ integers that meet the given conditions.

~lpieleanu (Minor editing)
~ rollover2020 (extremely minor editing)