2019 AMC 10B Problems/Problem 19

Revision as of 16:49, 14 February 2019 by Greersc (talk | contribs) (Solution)

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 $S^2$, which is $2^{10}\cdot 5^{10}$, so the answer is $121 - 4 = 117$.

Solution by greersc. (Edited by AZAZ12345 and then by greersc once again)

See Also

2019 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 18
Followed by
Problem 20
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All AMC 10 Problems and Solutions

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