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 $S?$

$\textbf{(A) }98\qquad\textbf{(B) }100\qquad\textbf{(C) }117\qquad\textbf{(D) }119\qquad\textbf{(E) }121$

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

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 $5^{10}$ . 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 $\boxed{\textbf{(C) } 117}$ .

-scrabbler94

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

Page

Toolbox

Search

2019 AMC 12B Problems/Problem 14

Problem

Solution

See Also