If AB and CA is a 2 digit number, and AB X 4 = CA, find A+B+C

The answer is 14.

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