2019 AMC 10A Problem 2

The hundreds digit is the third digit from the right. So the question is really about the last three digits of $20! - 15!$. A number ends in $000$ exactly when it is a multiple of $1000 = 2^3 \cdot 5^3$.

1. Count the 5s in $15!$. The product $1 \cdot 2 \cdot 3 \cdots 15$ contains $5$, $10 = 2 \cdot 5$ and $15 = 3 \cdot 5$. That is three factors of $5$.
2. Count the 2s in $15!$. There are many: $2$, $4$, $6$, $8$, and so on. Certainly at least three.
3. So $15!$ contains $2^3 \cdot 5^3 = 1000$ as a factor. It ends in $000$.
4. $20!$ as well. $20! = 15! \cdot 16 \cdot 17 \cdot 18 \cdot 19 \cdot 20$, so $20!$ is a multiple of $15!$, and therefore also a multiple of $1000$.
5. Subtract. If $20! = 1000a$ and $15! = 1000b$, then $20! - 15! = 1000(a - b)$, again a multiple of $1000$. It ends in $000$.

The last three digits are all $0$, so the hundreds digit is $0$.

The answer is $\boxed{\textbf{(A)}\ 0}$.

A question about one of the last few digits of a huge number is really a question about what it is divisible by: the last three digits are the remainder when you divide by $1000$. Factorials pick up a factor of $10$ for every multiple of $5$ they pass, so from $15!$ onward every factorial already ends in at least three zeros.

The product $15! = 1 \cdot 2 \cdots 15$ contains the factors $5$, $10$ and $15$, which supply $5^3$, and it contains far more than three factors of $2$. So $15!$ is divisible by $2^3 \cdot 5^3 = 1000$. Since $20!$ is a multiple of $15!$, it is divisible by $1000$ as well.

A difference of two multiples of $1000$ is a multiple of $1000$, so $20! - 15!$ ends in the digits $000$. Its hundreds digit is $0$.

The answer is $\boxed{\textbf{(A)}\ 0}$.