Question

1287a45b where the digits represented by a and b are different, is divisible by 18. For this 8-digit number to be the smallest, what digit must b be?

Answer 1

For the 8-digit number to be divisible by 18 it must be even and divisible by 9. The digits of all numbers divisible by 9 sum to a number divisible by 9. The sum of the digits in the given 8-digit number is 27, which is divisible by 9, so a+b must also be divisible by 9, the smallest number being 18. So a=1 and b=8 and they will produce the smallest 8-digit number: 1287a45b=12871458.