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Stella [2.4K]
2 years ago
10

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

Mathematics
1 answer:
musickatia [10]2 years ago
6 0

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}

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