Factorial addition

None of the digits can be $> 5$ because $7! = 5040 > 999$ and $6! = 720 > 700$ . Thus, our digits are bound to $\left[1, 5\right]$ . Observe that $4! + 3! + 2! = 32 < 100$ , so one of the digits must be $5$ . Also note that $1! + 2! + 3! + 4! + 5! = 153$ , meaning that $100 \leq \overline{ABC} \leq 153$ , so $A$ must equal $1$ . Because $A, B, C$ are distinct digits, our remaining digit is bound to $\left[2, 4\right]$ , making it easier to check for it. Let's examine the remaining cases: $\overline{1B5}$ or $\overline{15C}$ .

Case 1: $\overline{ABC} = \overline{15C}$

$\displaystyle \begin{aligned} 1! + 5! + C! & = \overline{15C} \\ 121 + C! & = 150 + C \\ \implies C! - C & = 29 \end{aligned}$

There exist no solutions $C \in \left[2, 4\right]$ for the above equation, so we must have $\overline{ABC} = \overline{1B5}$ .

Case 2: $\overline{ABC} = \overline{1B5}$

$\displaystyle \begin{aligned} 1 + B! + 5! & = \overline{1B5} \\ 121 + B! & = 100 + 10B + 5 \\ \implies 10B - B! & = 16 \\ \implies B & = 4 \end{aligned}$

We have $\overline{ABC} = 145$ , and verify that, indeed, $1! + 4! + 5! = 145$ .

Clearly none of A, B, or C can be 7 or higher, as 7! is a 4 digit number. If 6 were used, 6! =720, so we would need a 7 or higher as well. So the maximum possible is 5. If 5 is not used, the maximum sum is 4! + 3! + 2! = 32, which is not a long enough number. So one of the factorials used must be 5! = 120. Now we need the other two factorials to produce a 5 in the final number. So we need either to reach a total in the 30's (possible only with 4! + 3! = 30, but then 150 doesn't consist of the digits 5, 4, and 3) or the final digit must sum to 5 (possible only with 4! + 1! = 25, so the final total is 120 + 25 = 145.) This works, because 145 = 1! + 4! + 5!