A B and C

This solution is longer than necessary, but I want to establish a general process for this type of question, as well as to determine the uniqueness of the answer.

We require that $4*(10A + B) = 10C + A \Longrightarrow 39A + 4B = 10C$ , where $A,B,C$ are integers between $1$ and $9$ inclusive.

The maximum possible value for $10C$ is then $90$ , which means that, since $39A \gt 90$ for $A \ge 3$ , we must have either $A = 1$ or $A = 2$ . But since $10C$ is always even, we will require that $39A + 4B$ be even as well, which will only occurs for $A = 2$ .

So now we have that $39*2 + 4B = 10C \Longrightarrow 39 + 2B = 5C$ . Since $39 + 2B$ will always be odd and $\gt 40$ , we require the same for $5C$ , which is the case for only $C = 9$ . This requires that $39 + 2B = 45 \Longrightarrow B = 3$ .

Thus $A + B + C = 2 + 3 + 9 = \boxed{14}$ .

AB X 4 = CA. So, A = last digit of B X 4. If B = 1, then A = 4 => C=16(impossible). If B = 2, then A = 8 => C=32(impossible). If B = 3, then A = 12 => A = 2(carry 1) => C=9(possible). So, A = 2 , B=3 , C=9. => A + B + C = 14. Try the values with B as it is an independent variable.

$25 \times 4 = 100$

So, $10 \leq \overline{AB} \leq 24$

Now, $A = {1, 2}$ and $B = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}$

We know, $(even/odd) \times 4 = even$

Then, $\overline{CA}$ is a even number. Where the Last Digit $A$ will be Even.

Thus we get, $A = 2$ and $B = {0, 1, 2, 3, 4}$

Here, Only $B = 3$ satisfy the matter, where $B \times 4 = ...2$

So, $\overline{AB} = 23$

$23 \times 4 = 92$

So, $C = 9$

At last, $A + B + C = 2 + 3 + 9 = \boxed{14}$

It is called so Number Theory is awesome...

If anyone don't realize my speeches, then I must be sorry of that... :)

I just took the example 23(4)=92. Therefore, A=2,B=3 and C=9 Therefore, A+B+C=14