2017 AMC 10B Problem 16

A number can contain one $0$ or several, which makes direct counting messy. It is much easier to count the numbers with no $0$ and subtract from $2017$.

A number with no $0$ uses only the digits $1$ to $9$, so each place has $9$ choices. Count by the number of digits:

1. One-digit numbers ($1$ to $9$): $9$.
2. Two-digit numbers: $9 \cdot 9 = 81$.
3. Three-digit numbers: $9 \cdot 9 \cdot 9 = 729$.
4. Four-digit numbers from $1000$ to $1999$: the first digit is $1$, and the other three places have $9$ choices each: $729$.
5. Four-digit numbers from $2000$ to $2017$: every one has a $0$ in the hundreds place, so none of them count here.
6. Numbers with no $0$: $9 + 81 + 729 + 729 = 1548$.
7. Numbers with at least one $0$: $2017 - 1548 = 469$.

The answer is $\boxed{\textbf{(A)}\ 469}$.

"At least one" is often hard to count directly because of the many ways it can happen; "none" is usually a simple multiplication. Splitting by the number of digits keeps the first digit from being $0$, and the odd little tail $2000$ to $2017$ is easiest to check by hand.

Count the complement: integers from $1$ to $2017$ with no digit equal to $0$. Each digit can then be any of $1$ through $9$.

• One-digit numbers: $9$.
• Two-digit numbers: $9 \cdot 9 = 81$.
• Three-digit numbers: $9^3 = 729$.
• Four-digit numbers from $1000$ to $1999$: leading digit $1$, then $9^3 = 729$ choices for the rest.
• Four-digit numbers from $2000$ to $2017$: every one has a $0$ in the hundreds place, so $0$ zero-free numbers.

Zero-free total: $9 + 81 + 729 + 729 = 1548$.

Numbers containing a $0$: $2017 - 1548 = 469$.

The answer is $\boxed{\textbf{(A)}\ 469}$.