2019 AMC 12B Problems/Problem 14

Revision as of 13:15, 14 February 2019 by Robinaldabanx (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 $S?$

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 cases, and therefore the answer is $121-2 = \boxed{D}$

- Robin's solution

See Also

2019 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
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All AMC 12 Problems and Solutions
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