# 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 1

The ones digit, for all numbers divisible by 5, must be either $0$ or $5$. However, from the restriction in the problem, it must be even, giving us exactly one choice ($0$) for this digit. For the middle two digits, we may choose any even integer from $[0, 8]$, meaning that we have $5$ total options. For the first digit, we follow similar intuition but realize that it cannot be $0$, hence giving us 4 possibilities. Therefore, using the multiplication rule, we get $4\times 5 \times 5 \times 1 = \boxed{\textbf{(B) } 100}$. ~ciceronii swrebby

## Solution 2

The ones digit, for all the numbers that have to divisible be 5, must be a $0$ or a $5$. Since the problem states that we can only use even digits, the last digit must be $0$. From there, there are no other restrictions since the divisibility rule for 5 states that the last digit must be a $0$ or a $5$. So there are $4$ even digit options for the first number then $5$ for the middle 2. So when we have to do $4 \cdot 5 \cdot 5 \cdot 1 = \boxed{\textbf{(B) } 100}$. ~bobthefam

## Solution 3

This is basically 4*5*5 which is 100

~IceMatrix

~savannahsolver