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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; $1=1!$ $2=2!$ $145= 1!+4!+5!$ So what is the fourth self-factorial number? How about fifth? |
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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
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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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Its
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