Question

What is the smallest multiple of 18 of the form 2A945B, where A and B are digits?

Answer 1

Answer:

$219456$

Step-by-step explanation:

It must end in $0,2,4,6,8$ (divisible by 2 rule) and the digits must sum to a multiple of 9(divisible by 9 rule).

The digit sum is $2+A+9+4+5+B=20+A+B$.

We want to make $A$ as low as possible because it's a higher digit than $B$.

If $A$ is as low as possible, we have that the sum is $20+0+7=27$. Uh-oh. $B$ has to be even! So $A=0$ doesn't work.

If $A=1$, we have that the sum is $20+1+6=27$. Yay!

So, the answer is $\boxed{219456}$