2019 AMC 10B Problems/Problem 19

Problem

Let $S$ be the set of all positive integer divisors of $100,000.$ How many numbers are the product of two distinct elements of $S?$

$\textbf{(A) }98\qquad\textbf{(B) }100\qquad\textbf{(C) }117\qquad\textbf{(D) }119\qquad\textbf{(E) }121$

Solution

To find the number of numbers that are the product of two distinct elements of $S$ , we first square $S$ and factor it. Factoring, we find $S^2 = 2^{10} \cdot 5^{10}$ . Therefore, $S^2$ has $(10 + 1)(10 + 1) = 121$ distinct factors. Each of these can be achieved by multiplying two factors of $S$ . However, the factors must be distinct, so we eliminate $1$ and $S^2$ , so the answer is $121 - 1 - 1 = 119$ .

Solution by greersc.

2019 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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2019 AMC 10B Problems/Problem 19

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