A B
+ 3 7
______
6 B_
We need B x 3 = B
Since 5 x 3 = 15
∴ Possible value of B can be 5
Also 0 x 3 = 0, i.e. B = 0 can be another possible value
∵ A x 3 = A + 0 = A
∴ Possible value of A = 5
30, B must be
5 0
x 3
______
_1 5 0_
Thus, = 0
∴ A = 5, B = 0 and C = 1