Self-factorial number is the number where its digits' factorials summation is equal to the number itself. But there are only a few amount of them. For example;



$145= 1!+4!+5!$

So what is the fourth self-factorial number? How about fifth?

  • $\begingroup$ Are we using base ten? $\endgroup$ – Ankoganit Dec 24 '16 at 13:47
  • $\begingroup$ @Ankoganit of course :) $\endgroup$ – Oray Dec 24 '16 at 13:47

A self-factorial number can have no more than 7 digits. Proof: Suppose it has $n$ digits. Then the sum of the factorials of the digits is at most $$ 9! \cdot n = 362880n $$ whereas the number itself is at least $$ 10^{n-1}. $$ For $n \ge 8$, $10^{n-1} > 362880n$ so these two cannot be equal.

So it remains that we check all numbers with up to $7$ digits. A brute-force search checking $1$ to $9999999$ finds that the only solutions are

1, 2, 145, 40585.

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  • $\begingroup$ i like this one, thanks for showing that it is not possible to have more than 4 numbers. $\endgroup$ – Oray Dec 24 '16 at 16:17


40585. They are factorians-A014080. And no there is no fifth. Reference

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  • $\begingroup$ lol I did not know it was defined :) $\endgroup$ – Oray Dec 24 '16 at 14:33

I wrote a Java program to find a solution by brute-force. I found


as a fourth number.

The fifth number should have a ton of digits. I'm still searching.

EDIT: it looks like there is not a fifth one.

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  • 9
    $\begingroup$ It's a Java program. It'll take it some time to count to 5. $\endgroup$ – UTF-8 Dec 24 '16 at 18:36

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