2019 AMC 10A Problem 2

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 $\boxed{\textbf{(A)}\ 0}$.

Questions about a specific digit near the end of a huge number are divisibility questions: the last three digits are the remainder mod $1000$. Factorials pick up factors of $10$ quickly (one for every multiple of $5$ up to the end), so $n!$ for $n \ge 15$ is a multiple of $1000$ and the hundreds digit is automatically $0$.

• Trying to compute the factorials or their last digits by hand instead of noticing both are multiples of 1000.
• Counting only the visible factor $10$ in $15!$ and concluding it has just one trailing zero; the factors $5$ and $15$ pair with even numbers to give two more.