2019 AMC 10B Problem 19

$100{,}000 = 10^5 = 2^5 \cdot 5^5$, so $S$ consists of the numbers $2^a 5^b$ with $0 \le a, b \le 5$.

A product of two elements of $S$ has the form $2^x 5^y$ with $0 \le x, y \le 10$. There are $11 \cdot 11 = 121$ such numbers, and each one arises: split $x = a_1 + a_2$ and $y = b_1 + b_2$ with all parts in $\{0, \dots, 5\}$.

The question requires the two divisors to be distinct. An exponent $x$ can be split into two different parts unless $x = 0$ (only $0 + 0$) or $x = 10$ (only $5 + 5$). So $2^x 5^y$ is a product of two distinct divisors as long as at least one of $x, y$ lies strictly between $0$ and $10$; the exceptions are the four "corners" $x, y \in \{0, 10\}$:
$$1 = 1 \cdot 1,\quad 2^{10} = 2^5 \cdot 2^5,\quad 5^{10} = 5^5 \cdot 5^5,\quad 10^{10} = 10^5 \cdot 10^5.$$

Hence $121 - 4 = 117$ numbers qualify. The answer is $\boxed{\textbf{(C)}\ 117}$.

Multiplying divisors adds exponents, so the set of all products is a grid of exponent pairs, easy to count. The "distinct" condition is then a small correction: a product is forced to use equal factors only when both exponents are at an extreme. Count the whole grid, then subtract the genuinely degenerate points.

• Answering 121 by forgetting the word 'distinct', or subtracting only the two obvious cases 1 and 10^10 to get 119.
• Excluding every product of the form $d \cdot d$ (all $36$ squares), forgetting that most perfect-square products can also be written with two different divisors.