# 2019 AMC 10B Problems/Problem 19

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The following problem is from both the 2019 AMC 10B #19 and 2019 AMC 12B #14, so both problems redirect to this page.

## 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

The prime factorization of $100,000$ is $2^5 \cdot 5^5$. Thus, we choose two numbers $2^a5^b$ and $2^c5^d$ where $0 \le a,b,c,d \le 5$ and $(a,b) \neq (c,d)$, whose product is $2^{a+c}5^{b+d}$, where $0 \le a+c \le 10$ and $0 \le b+d \le 10$.

Notice that this is similar to choosing a divisor of $100,000^2 = 2^{10}5^{10}$, which has $(10+1)(10+1) = 121$ divisors. However, some of the divisors of $2^{10}5^{10}$ cannot be written as a product of two distinct divisors of $2^5 \cdot 5^5$, namely: $1 = 2^05^0$, $2^{10}5^{10}$, $2^{10}$, and $5^{10}$. The last two cannot be written because the maximum factor of $100,000$ containing only $2$s or $5$s (and not both) is only $2^5$ or $5^5$. Since the factors chosen must be distinct, the last two numbers cannot be so written because they would require $2^5 \cdot 2^5$ or $5^5 \cdot 5^5$. The first two would require $1 \cdot 1$ and $2^{5}5^{5} \cdot 2^{5}5^{5}$, respectively. This gives $121-4 = 117$ candidate numbers. It is not too hard to show that every number of the form $2^p5^q$, where $0 \le p, q \le 10$, and $p,q$ are not both $0$ or $10$, can be written as a product of two distinct elements in $S$. Hence the answer is $\boxed{\textbf{(C) } 117}$.

## Video Solution by OmegaLearn

~ pi_is_3.14

 2019 AMC 10B (Problems • Answer Key • Resources) Preceded byProblem 18 Followed byProblem 20 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions
 2019 AMC 12B (Problems • Answer Key • Resources) Preceded byProblem 13 Followed byProblem 15 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 12 Problems and Solutions