2019 AMC 10B Problem 19

First, note that $100,000=2^5\cdot5^5.$ Therefore, any element of $S$ must be of the form $2^a5^b$ with $0 \leq a,b \leq 5.$ Suppose I have distinct $x,y \in S$ with $$x = 2^a5^b,$$$$y=2^c5^d.$$ Then, $$xy = 2^{a+c}5^{b+d}.$$ Thus, $$0 \leq a+c,b+d \leq 10.$$ This means that there are $$(10+1)(10+1)=121$$ possible exponent pairs for a product. However, some products can arise only when $x=y$, namely when $(a,b)=(c,d)$. If $a+c=0,$ then $a=0,c=0$ must be true. If $a+c=10,$ then $a=5,c=5$ must be true. With any other value of $a+c,$ we can have $a \neq c.$ Similar structure holds for $b+d.$ Thus, if $$a+c,b+d \in \{0,10\},$$ then $$(a,b)=(c,d),$$ thus making $x=y.$ This means we have to eliminate $4$ choices, leaving $$121-4=117.$$ Thus, the answer is C .