It's a very old AMC/AIME problem. 145 is the only 3-digit factorian

This should be moved to MSM! It's really easy for HSO.

ZeusDM wrote: 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. If exists a number $N=\overline{abc}$ such, then $(a,b,c) \le 5$. This is becuase $6!=720<1000$ but at least you will need an $a \ge 7$ and $7!=5040$, so that's not possible. $a \le b \le c \le 5$. This is becuase If $a=5$, then $b=c=5$ but still not a value larger than $500$, in fact $500>5!+5!+5!=360$. If $b=5$, then $a \le 2$. Case 1: If $a=2$. $N=\overline{25c}$ then $c=5$ at least but still no larger than requied. $250>2!+5!+5!=244$. Case 2: If $a=1$. $N=\overline{15c}$ then no such $c$ becuase $150>1!+5!+4!=145$. Then $c=5$ and $a,b$ are least or equal to $c$. $N=\overline{ab5}$ and we saw there is no solutions with $b=5$ then $b \le 4$ and $a=1$. $N=\overline{1b5}$ here we bash: $105+10b=121+b! \implies 10b-b!=16$ we can see that the only number such is $b=4$. Then $N=145$ is the only solution. Thus we are done