AMC10 2019 B

Answer (C): Note that 100,000 = $2^5 \cdot 5^5$. This implies that for a number to be a product of two elements in $S$ it must be of the form $2^a \cdot 5^b$ with $0 \le a \le 10$ and $0 \le b \le 10$. The corresponding product for the remainder of this solution will be denoted $(a, b)$. Note that the pairs $(0,0)$, $(0,10)$, $(10,0)$, and $(10,10)$ cannot be obtained as the product of two distinct elements of $S$; these products can be obtained only as $1 \cdot 1 = 1$, $5^5 \cdot 5^5 = 5^{10}$, $2^5 \cdot 2^5 = 2^{10}$, and $10^5 \cdot 10^5 = 10^{10}$, respectively. This gives at most $11 \cdot 11 - 4 = 117$ possible products. To see that all these pairs can be achieved, consider four cases: If $0 \le a \le 5$ and $0 \le b \le 5$, other than $(0,0)$, then $(a, b)$ can be achieved with the divisors $1$ and $2^a \cdot 5^b$. If $6 \le a \le 10$ and $0 \le b \le 5$, other than $(10,0)$, then $(a, b)$ can be achieved with the divisors $2^5$ and $2^{a-5} \cdot 5^b$. If $0 \le a \le 5$ and $6 \le b \le 10$, other than $(0,10)$, then $(a, b)$ can be achieved with the divisors $5^5$ and $2^a \cdot 5^{b-5}$. Finally, if $6 \le a \le 10$ and $6 \le b \le 10$, other than $(10,10)$, then $(a, b)$ can be achieved with the divisors $2^5 \cdot 5^5$ and $2^{a-5} \cdot 5^{b-5}$.