<s>[INST] Factorial Digits Sum 
A natural number is a factorion if it is the sum of the factorials of
  each of its decimal digits. For example, $145$ is a factorion because
  $145 = 1! + 4! + 5!$. Find every $3$-digit number which is a factorion.

My solution: 
We can obviously only have factorials $1$–$6$. 


*

*If we have a $6$ factorial, our number would need to start with $6$ to be the biggest possible, but that is still too small. Doesn't work. 

*If we have a $5$ factorial, since that's $120$, we need our number to start with $1$. $5!+1!=121$. We need a $1\_\_$. If after the $1$ we have a $5$, we can do the casework. 


*

*$151$ does not work. 

*Neither does $152$. 

*Neither does $153$.

*Neither does $154$.


*But we did the computations for $151$, $152$, $153$, and $154$ to find this, and we notice that $154$'s answer is $145$. So, $145$ works, and we quickly see there are no others, so only $145$ works.


How can we generalize this?
 [/INST] Brainstorming:  
As you pointed out.  If one of the digits of $N$ is $7$ then $7! > 1000> N$ which is impossible.
If one of the digits is $6$ then $N > 6! = 720$ so one of the digits is at least $7$ which we just showed was impossible.
Also if all three digits, call them $a,b,c$, are all $< 5$ then $a!, b!,c! \le 4! =24$ and $N = a! + b!+c! \le 3*24 = 72 < 100$ so that's impossible.  So there must be at least one digit that is equal to $5$.
So one of the digits is $5$. Then if the other two are $a,b$ then $a,b \le 5$ and $N = a! + b! + 5! \le 5! + 5! + 5! = 360$.  So the first digit is $3$ or less.
So $N = a! + b! +5! \le 3! + 5! + 5! = 246$ so the first digit is $2$ or less.
If the first digit, let's assume the first digit is $a$ is equal $2$ then $200 < N = 2! + b! + 5! = b! + 122$ so $b! > 78$ so $b >4$ but $b \le 5$ so $b=5$ and $N = 255$ but $2! + 5! + 5! = 242$ so $255$ is not factorial.  So the first digit is less than $2$.
But $N > 100$ so the first digit is $1$.
So $N = 1! + b! + 5! = 126 +b!$.  Where $b = 0,1,2,3,4,5$ and I guess at this point we can do trial and error. 
But we have $100 < N < 200$ and $N = 126+b!$ so $b!  < 74$ for $b \le 4$.  So $N=126 +b! \le 126 + 24 = 150$.  And $1! + 5! + 0! = 122\ne 150$ so $150$ is not factorial so $N \le 149$.  but as one of the digits is $5$ that must be the last digit and $N = 100 + 10b + 5$.
So $N = 105 + 10b = 1! + b! + 5! = 121 + b!$ so $10b = b! + 16$.  And we know $N>121$ so $b \ge 2$ and $b \le 4$ and $b! + 16 \equiv 0 \pmod {10}$ and.... how much further can I go without saying, look, $b$ has got to be $4$; that's the only option?
Well a little further.... just to be masochistic.  $10b = 20,30$ or $40$ so $b! = 4,14$ or $24$ and $b = 2,3,4$ so either $2! = 4$, $3!=14$ or $4! = 24$ and ....
That's as far as I can go.
$b =4$ and $N = 100 + 40 + 5 = 1! + 4! +5!$.
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