Since, you forgot the options, I can help only a little.
Multiples of 3 have the property that the sum of their digits has to be divisible by 3.
so for the first number it means that 7+4+A+5+2+B+1 must be divisible by 3.
let's simplify this:
7+4+A+5+2+B+1=11+A+7+B+1=19+A+B
we can actually from now on think in terms of :A+B+1 is divisibe by 3, since 19 =18+1 and 19
and the same for the second number
3+2+6+A+B+4+C=15+A+B+C
15 is divisible by 3, so we can just remember that
A+B+C is divisible by 3.
A+B+1 is divisive by 3
C must be of the form 3x+2 then (since A+B+1 is divisible by 3_: it can be either 2 (3*0+2), 5, or 8 - since the next possibility, 11 is bigger than 10, so it cannot be a digit in this number!
