2019 AMC 12B Problem 14

Since $100{,}000=2^5\cdot5^5,$ every divisor is $2^a5^b$ with $0\le a,b\le5.$ A product of two divisors is $2^x5^y$ with $0\le x,y\le10,$ and every such pair $(x,y)$ is attainable, giving $11\cdot11=121$ values. We need two distinct divisors. A value $2^x5^y$ is forced to be a divisor times itself only when both $x$ and $y$ have a unique split, which happens exactly when $x,y\in\{0,10\}.$ Those $4$ corner values ($1,\ 2^{10},\ 5^{10},\ 2^{10}5^{10}$) cannot use two distinct divisors. The count is $121-4=117.$ Thus, C is the correct answer.