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

User Herb
by
3.1k points

1 Answer

21 votes
21 votes

Answer:


219456

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}

User Brandon Moretz
by
2.8k points