The multiplication of 3 and B gives a number whose ones digit is B again

Therefore, B must be 0 or 5

Let B is 5

Multiplication of first step = 3 × 5 = 15

Now,

1 will be a carry for the next step

3 × A + 1 = CA

This is not valid for any value of A

Hence, B must be 0 only.

If B = 0, then there won’t be any carry for the next step

We will get, 3 × A = CA

That is, the one's digit of 3 × A should be A

This is possible when A = 5 or 0

But, 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