The multiplication of 3 and B gives a number whose ones digit is B again.
Hence, B must be 0 or 5.
Let B is 5.
Multiplication of first step = 3 × 5 = 15
1 will be a carry for the next step.
We have, 3 × A + 1 = CA
This is not possible for any value of A.
Hence, B must be 0 only. If B = 0, then there will be no carry for the next step. We should obtain, 3 × A = CA
That is, the one’s digit of 3 × A should be A. This is possible when A = 5 or 0.
However, A cannot be 0 as AB is a two-digit number.
Therefore, A must be 5 only. The multiplication is as follows.
Hence, the values of A, B, and C are 5, 0, and 1 respectively.