Numbers Made of Their Own Factorials

The number $145$ has a curious property: add the factorials of its digits and you get the number back, $$1! + 4! + 5! = 1 + 24 + 120 = 145.$$ Find every number, beyond the trivial $1$ and $2$, equal to the sum of the factorials of its digits.

Solution

There are just two more, and the complete list is $$1, \quad 2, \quad 145, \quad 40585.$$ The last checks out as $4! + 0! + 5! + 8! + 5! = 24 + 1 + 120 + 40320 + 120 = 40585$ (recall $0! = 1$).

The reason the hunt ends is a race between two quantities. A number with $d$ digits is at least $10^{d-1}$. The most its digit factorials can ever total is $d \times 9! = 362880\,d$, since $9!$ is the biggest single-digit factorial. For $d = 8$ the number is already at least $10^{7} = 10{,}000{,}000$, while its digit factorials can reach at most $8 \times 362880 = 2{,}903{,}040$, which is smaller. From eight digits on the number always outruns the sum, so no large example can exist, and a finite check settles the rest. Such numbers are called factorions, and in base ten these four are all there are.