# Difference between revisions of "2020 AMC 10A Problems/Problem 6"

The following problem is from both the 2020 AMC 12A #4 and 2020 AMC 10A #6, so both problems redirect to this page.

## Problem

How many $4$-digit positive integers (that is, integers between $1000$ and $9999$, inclusive) having only even digits are divisible by $5?$ $\textbf{(A) } 80 \qquad \textbf{(B) } 100 \qquad \textbf{(C) } 125 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 500$

## Solution

The units digit, for all numbers divisible by 5, must be either $0$ or $5$. However, since all digits are even, the units digit must be $0$. The middle two digits can be 0, 2, 4, 6, or 8, giving 5 choices for each. The first (thousands) digit can be 2, 4, 6, or 8, giving 4 choices. Therefore we get $4\times 5 \times 5 \times 1 = \boxed{\textbf{(B) } 100}$ possible 4-digit integers.

## Video Solution 1

Education, the Study of Everything

~IceMatrix

## Video Solution 3

~savannahsolver

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