The sum of the positive divisors of a positive integer of the form $2^i3^j$ is equal to $600$. What is $i + j$?
I can rewrite this as $(2^{i+1} - 1)(3^{j+1} - 1) = 1200$. Now I need to factor 1200 into two factors that are one less than powers of 2 and 3, respectively. I see that 1200 has a lot of factors of 2, so I start by looking at the highest power of 2 that is one less than a factor of 1200. That would be $2^4 - 1 = 15$, which is a factor of 1200. So I can try $2^{i+1} - 1 = 15$, which implies $i = 3$. Then I need to find $j$ such that $3^{j+1} - 1 = 1200 / 15 = 80$. I see that $3^4 - 1 = 80$, so $j = 3$ as well. Therefore, $i + j = 3 + 3 = 6$.

