For positive integer $n$ such that $n < 10000$, the number $n+2005$ has exactly 21 positive factors. What is the sum of all the possible values of $n$?
Given that $n + 2005$ has $21$ positive factors, $n + 2005$ can be written as $p_1^2 \cdot p_2^6$ because $(2+1)(6+1) = 21$. Also, $n + 2005$ must be greater than $2005$ and less than $12005$ because $0 < n < 10000$. My plan was to sum up the possible values of $n+2005$ and then subtract $2005$ times the number of possible values. But this seems to be quite hard since for example, let $p_2 = 2$, then $p_2^6$ = 64, and then there are many squares of prime numbers that can be $p_1$. When I searched up "Sum of prime squares", it showed An easy way to count sum of squares of first x prime numbers? as one of the results. It says that there is no closed form on the accepted answer, or no exact expression. So I have to brute out the squares and add them one by one, which isn't a good idea. Is there an easier way (answer is probably yes)? What is it?
