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2017 AMC 10B Problem 16

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2017 AMC 10B Problem 16AMC 10 Step by Step

Name: Date: September 13, 2026

Bases & Digits · difficulty 3 of 5 · about 4 min · position P16-20

How many of the base-ten numerals for the positive integers less than or equal to $2017$ contain the digit $0$ ?

A)
$469$
B)
$471$
C)
$475$
D)
$478$
E)
$481$

Answer

Key insight

Count numbers with no zero digit (9, 81, 729 for 1 to 3 digits, 729 for 1000-1999, none for 2000-2017) and subtract from 2017.

Solution

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 $\boxed{\textbf{(A)}\ 469}$ .

Why this works

"Contains at least one $0$ " is hard to count directly because a number may contain several zeros; "contains no $0$ " is a clean product of independent digit choices. Splitting by digit length handles the leading-digit restriction, and the awkward tail $2000$ to $2017$ is checked separately rather than forced into a formula.

The trap

Forgetting that every number from 2000 to 2017 contains a 0, or allowing 0 as a leading digit in the no-zero count.

Common mistakes

Forgetting that every number from 2000 to 2017 contains a 0, or allowing 0 as a leading digit in the no-zero count.
Counting directly by position of the zero and double counting numbers such as $100$ or $1001$ that have two zeros.

Techniques

Count the complement and subtract from the total