2019 AMC 12B Problems/Problem 14

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 $S?$

Solution

First, find the prime factorization of $100,000$ . It is $2^5 \cdot 5^5$ . Thus, any factor will have the pattern $2^m \cdot 5^n$ , where $m, n < 5$ . Multiplying this by another factor with the same pattern $2^k \cdot 5^l$ gets us $2^m+k \cdot 5^n+l$ . It initially seems like we have $11$ options for the power of $2$ and $11$ options for the power of $5$ , giving us a total of $121$ choices. However, note that the factors must be distinct. If they are distinct, we cannot have $1$ (as it is only formed by $1 \cdot 1$ ), or $2^{10} \cdot 5^{10}$ (as it is only formed by $100,000 \cdot 100,000$ ). These are the only two cases where the distinction rule forces us to eliminate case, and therefore the answer is $121-2 = \boxed{D}$

- Robin's Solution

2019 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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2019 AMC 12B Problems/Problem 14

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