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.
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?