Question 171070
Starting with 12A3B.
To be divisible by 4, the last two digits (3B) must be divisible by 4, so B must be either 2 (32 = 4X8) or 6 (36 = 9X4)
Let's assume that B is 2, so then we have:
12A32 Now for this to be divisible by 9, the sum of the digits must be divisible by 9 so that 1+2+A+3+2 is a multiple of 9.
Add the digits: 1+2+3+2+A = 8+A So A must be 1 because 8+1 = 9
Let's check: 
{{{12132/4 = 3033}}} So it's divisible by 4. Now try 9
{{{12132/9 = 1349}}} and it's also divisible by 9. 
A = 1 and B = 2