# Difference between revisions of "2006 AMC 10A Problems/Problem 21"

## Problem

How many four-digit positive integers have at least one digit that is a 2 or a 3?

$\mathrm{(A) \ } 2439\qquad\mathrm{(B) \ } 4096\qquad\mathrm{(C) \ } 4903\qquad\mathrm{(D) \ } 4904\qquad\mathrm{(E) \ } 5416\qquad$

## Solution

Since we are asked for the number of positive 4-digit integers with at least 2 or 3 in it, we can find this by finding the total number of 4-digit integers and subtracting off those which do not have any 2s or 3s as digits.

The total number of 4-digit integers is $9 \cdot 10 \cdot 10 \cdot 10 = 9000$, since we have 10 choices for each digit except the first (which can't be 0).

Similarly, the total number of 4-digit integers without any 2 or 3 is $7 \cdot 8 \cdot 8 \cdot 8 =\boxed{3584}$.

Therefore, the total number of positive 4-digit integers that have at least one 2 or 3 in their decimal representation is $9000-3584=5416 \Longrightarrow \mathrm{(E)}$