$\overline{xyz} =x!+y!+z!$ how to find all of 3-digit integer number $\overline{xyz}$  such that:
$\overline{xyz} =x!+y!+z!$
 A: Here is how a human can solve this.


*

*No digit can be $7$, $8$ or $9$, because $7! > 1000$.

*No digit can be a $6$ either. Otherwise, $\overline{xyz}$ would be at least $720$, and its leading digit would be at least $7$, contradicting step 1.

*There must be at least one digit $5$. Otherwise the whole number would not be greater than $3 \cdot 4! < 100$, and it would have at most two digits. So, there is a $5$.

*There is exactly one digit $5$. There can't be three because number $555$ clearly doesn't fit. There can't be two either: otherwise $\overline{xyz}$ is somewhere between $5!+5!+0!$ and $5!+5!+4!$, which means that the leading digit is $2$. So the number is $255$, but it doesn't fit either. So, there is exactly one $5$.

*Since there is exactly one $5$, $\overline{xyz}$ is between $5!+0!+0!$ and $5!+4!+4!$. In any case the leading digit is $1$. So, the digits are $1$, $5$ and something from $0$ to $4$. Now there are only five options, which is OK to check by hand. It turns out that the digits are $1$, $4$ and $5$, and the answer is $145$.


UPDATE: hm, after reading the first lines of Michael Albanese's answer I realized that I completely ignored the existence of digit $0$. Fortunately, this doesn't affect the logic of my answer at all. I've updated the text above to get rid of this error.
A: Try ruby:
def fac(n); s = 1; n.times do |i| s*= i+1 end; s end
(1..999).select do |n| n == fac(n%10) + fac(n/10%10) + fac(n/100) end

Good luck ;-)
A: As I was partway through typing this answer, Dan Shved's answer appeared which is much better than mine, but I'll include what I have so far as I use some different ideas. I reduce the number of possibilities to $13$ (though of course, I could reduce this number even further using the reasons in Dan's answer).

\begin{align*}
0! &= 1\\
1! &= 1\\
2! &= 2\\
3! &= 6\\
4! &= 24\\
5! &= 120\\
6! &= 720
\end{align*}
As $7! = 5040$, each of $x, y,$ and $z$ is between $0$ and $6$. Furthermore, at most one of $x, y,$ and $z$ can be $6$ otherwise $x! + y! + z!$ is not a three digit number (too big). Similarly, at least one of $x, y,$ and $z$ is a $5$ or $6$ otherwise $x! + y! + z!$ is not a three digit number (too small). Also, $x$ can't be $0$. We are left with $229$ possibilities.
Note that $\overline{xyz}$ can't end with a $0$ or a $1$ because there is no possible $x$ and $y$ such that $x! + y! \equiv 9\ \textrm{mod}\ 10$.
If $\overline{xyz}$ ends with a $2$ or a $4$, we must have $\overline{xy} \in \{34, 43, 56, 65\}$
as these are the only choices with $x! + y! \equiv 0\ \textrm{mod}\ 10$.
If $\overline{xyz}$ ends with a $3$ we must have $\overline{xy} \in \{13, 31\}$  as these are the only choices with $x! + y! \equiv 7\ \textrm{mod}\ 10$.
If $\overline{xyz}$ ends with a $5$ we must have $\overline{xy} \in \{14, 40, 41\}$  as these are the only choices with $x! + y! \equiv 5\ \textrm{mod}\ 10$.
If $\overline{xyz}$ ends with a $6$ we must have $\overline{xy} \in \{24, 42\}$ as these are the only choices with $x! + y! \equiv 6\ \textrm{mod} 10$.
So we are left with $15$ possibilities: 
$$342, 432, 562, 652, 344, 434, 564, 654, 133, 313, 145, 405, 415, 246, 426.$$
Note that $\overline{xyz}$ is divisible by $\min\{x!, y!, z!\} = (\min\{x, y, z\})!$. This condition is automatically satisfied if the smallest digit is $0, 1,$ or $2$. However, this condition rules out both $564$ and $654$ as they're not divisible by $4!$.
