2019 AMC 10B Problem 19

Multiplying numbers adds their exponents, so write everything with exponents.

1. $100{,}000 = 10^5 = 2^5 \cdot 5^5$. Its divisors are the numbers $2^a \cdot 5^b$ where $a$ and $b$ are each one of $0, 1, 2, 3, 4, 5$. (For example $2^3 \cdot 5^1 = 40$.)
2. Multiply two divisors, $2^{a_1} 5^{b_1}$ and $2^{a_2} 5^{b_2}$. The product is $2^{a_1 + a_2} \cdot 5^{b_1 + b_2}$. Each total exponent is between $0$ and $10$.
3. Every pair of totals from $0$ to $10$ can be reached (for example, $7 = 5 + 2$). So the possible products are $2^x \cdot 5^y$ with $x$ and $y$ each from $0$ to $10$. That is $11 \cdot 11 = 121$ numbers.
4. Now use "distinct". Can we reach $2^x 5^y$ with two different divisors? The two divisors are the same only if $a_1 = a_2$ and $b_1 = b_2$. So we only need one exponent that can be split unevenly.
- $x = 0$ can only be split as $0 + 0$, and $x = 10$ only as $5 + 5$. Every $x$ from $1$ to $9$ can be split into two different parts (for example $4 = 3 + 1$, $9 = 5 + 4$).
- The same holds for $y$.
5. So $2^x 5^y$ fails only when both $x$ and $y$ are $0$ or $10$. Those four products are
$$1 = 1 \cdot 1, \qquad 2^{10} = 32 \cdot 32, \qquad 5^{10} = 5^5 \cdot 5^5, \qquad 10^{10} = 10^5 \cdot 10^5.$$
Each one can only be made by multiplying a divisor by itself.
6. The count is $121 - 4 = 117$.

The answer is $\boxed{\textbf{(C)}\ 117}$.

Divisors of a number like $2^5 \cdot 5^5$ correspond to pairs of exponents, and multiplying divisors adds those exponents. That turns the question into counting points on a grid. The word "distinct" only removes the corner points, where both exponents are forced to split evenly. Check such small exceptions carefully: all four corners count, not just $1$ and $10^{10}$.

$100{,}000 = 10^5 = 2^5 \cdot 5^5$, 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 $\boxed{\textbf{(C)}\ 117}$.