What is the smallest four-digit positive integer that is divisible by both 5 and 9 and has only even digits?
Solution 1

Suppose the number is abcd where $a, b, c$, and $d$ are digits.

Since the number is divisible by 5 , we must have that $d=0$ or $d=5$.

The digits are all even, which means $d \neq 5$, so $d=0$.

The smallest that $a$ can be is 2 since it must be even and greater than 0 (a four-digit number cannot have $a=0$ ). So we will try to find such a number with $a=2$.

In order to be divisible by 9 , we must have $a+b+c+d$ divisible by 9 . Substituting $a=2$ and $d=0$, this means $2+b+c+0=2+b+c$ is divisible by 9 .

Since $b$ and $c$ are even, $2+b+c$ is even, which means it cannot equal 9 . Thus, we will try to find $b$ and $c$ so that $2+b+c=18$, which is the smallest multiple of 9 that is greater than 9 .

This equation rearranges to $b+c=16$. Since $b$ and $c$ are even and satisfy $0 \leq b \leq 9$ and $0 \leq c \leq 9$, the only possibility is $b=c=8$.

## Solution 2

An integer is divisible by both 5 and 9 exactly when it is divisible by 45 .

Since we seek a number having only even digits, its last digit is one of $0,2,4,6$, or 8 , so the number itself is even.

Thus, we are looking for an even multiple of 45.

An even number is a multiple of 45 exactly when it is a multiple of 90 .

This means we are looking for the smallest 4-digit multiple of 90 that has only even digits. The smallest 4-digit multiple of 90 is 1080, but the first digit of this number is 1 , which is odd. Each of the next 10 multiples of 90 has a first digit equal to 1, so the number we seek must be larger than 2000 .

The four-digit multiples of 90 that have a thousands digit of 2 are

$$
2070,2160,2250,2340,2430,2520,2610,2700,2790,2880,2970
$$

and the only number in this list that has all even digits is 2880 .

Therefore, 2880 is the smallest 4-digit number that is a multiple of 5, a multiple of 9 , and has only even digits.