Question

What is the smallest multiple of 18 of the form 2A945B, where A and B are digits? Enter the full 6-digit number as your answer.

Answer 1

Number to be divisible by 18 , it must be divisible by 2 and 9

For number to divisible by 2 unit digit must be even .

For number to be divisible by 9 sum of digits must be divisible by 9.

Sum of digits = 2+A+9+4+5+B= A+B+20

So A+B must be equal to 7 for number to divisible by 9

7 = 1+6 , 2+5 , 3+4

So A=1 , B=6

Smallest even number = 219456