# Difference between revisions of "2017 AMC 10B Problems/Problem 16"

## Problem

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

$\textbf{(A)}\ 469\qquad\textbf{(B)}\ 471\qquad\textbf{(C)}\ 475\qquad\textbf{(D)}\ 478\qquad\textbf{(E)}\ 481$

## Solution

We can use complementary counting. There are $2017$ positive integers in total to consider, and there are $9$ one-digit integers, $9 \cdot 9 = 81$ two digit integers without a zero, $9 \cdot 9 \cdot 9$ three digit integers without a zero, and $9 \cdot 9 \cdot 9 = 1458$ three-digit integers without a zero. Therefore, the answer is $2017 - 9 - 81 - 729 - 729 = \boxed{\textbf{(A) }469}$.

## See Also

 2017 AMC 10B (Problems • Answer Key • Resources) Preceded byProblem 15 Followed byProblem 17 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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