Let $S$ be the set of all positive integer divisors of $100000$. How many numbers are the product of $2$ distinct elements of $S$?
factors of 100000
Let $100000 = 2^55^5$ be divisible by $2^a5^b$ and $2^b5^c$. Then $a, b, c, d$ are nonnegative integers less than $6$. Since $0 \le a + c, b + d \le 10$, there exist $11^2 = 121$ possible values for $(2^a5^b)(2^c5^d) = 2^{a + c}5^{b + d}$. $1, 2, 5, 2^{10}5^{10}$ are the only values that cannot equal the product of $2$ distinct factors of $100000$. As a result, the answer is $121 - 4 = \boxed{117}$.
Am I right? Any other solutions?