# Difference between revisions of "2019 AMC 10B Problems/Problem 19"

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 1

To find the number of numbers that are the product of two distinct elements of $S$, we first square $100,000$ and factor it. Factoring, we find $100,000^2 = 2^{10} \cdot 5^{10}$. Therefore, $100,000^2$ 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 $100,000^2$, as well as $2^{10}$ and $5^{10}$, so the answer is $121 - 4 = 117$. $2^{10}$ and $5^{10}$ do not work since the factors chosen must be distinct, and those require $2^5 \cdot 2^5$ or $5^5 \cdot 5^5$.

## Solution 2

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 $0 \le b+d \le 10$.

Consider $100000^2 = 2^{10}5^{10}$. The number of divisors is $(10+1)(10+1) = 121$. 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 can not be created because the maximum factor of 100,000 involving only 2s or 5s is only $2^5$ or $5^5$. Since the factors chosen must be distinct, the last two numbers cannot be created because those require $2^5 \cdot 2^5$ or $5^5 \cdot 5^5$. 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 $\boxed{\textbf{(C) } 117}$.