Find the number which is sum of the factorial of each of its digits.

Given that $A$ , $B$ , and $C$ are all positive, we can start drawing conclusions about the three numbers.

1) the maximum of $\{A,B,C\}$ is at most $5$ . Clearly the maximum can not be $7$ or more, as $7! = 5040$ , which is too big for our sum. Also, since $6! = 720$ , if $6$ is one of the three numbers, $A$ is at least $7$ , which again is not possible.

2) the maximum of $\{A,B,C\}$ is at least $5$ . Since $4! = 24$ , if the maximum of $\{A,B,C\}$ is less than $4$ , then $ABC = A! + B! + C! \leq 72$ , a contradiction.

3) It is not possible that $A \in \{4,5\},$ since $A! + B! + C! \leq 3 \times 5! = 360$ .

4) $A = 3$ is not possible, as that would imply $A! + B! + C! \leq 3! + 2*5! = 246 < 300$ , a contradiction.

5) Suppose $A = 2$ . Consider the possibility $\{B,C\} = \{4,5\}$ . But 2! + 4! + 5! = 146, which would contradict $A=2$ . Thus it must be the case that $B= C = 5$ . But $2! + 5! + 5! = 242$ , which does not satisfy the condition.

6) Thus $A = 1$ and either $B = 5$ or $C = 5$ . At this point we can simply run through the five possible sums:

• $1! + 1! + 5! = 122$
• $1! + 2! + 5! = 123$
• $1! + 3! + 5! = 127$
• $1! + 4! + 5! = 145$
• $1! + 5! + 5! = 241$

It is clear that the only solution is ABC = 145.