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I am trying to understand the upper bound on factorions (in base $10$). The Wikipedia page says:

"If $n$ is a natural number of $d$ digits that is a factorion, then $10^{d − 1} \le n \le 9!d$. This fails to hold for $d \ge 8$ thus $n$ has at most $7$ digits, and the first upper bound is $9,999,999$. But the maximum sum of factorials of digits for a $7$ digit number is $9!7 = 2,540,160$ establishing the second upper bound."

Please explain this to me in simple terms, precisely the first part $10^{d − 1} \le n \le 9!d$.

share|cite|improve this question… asks essentially the same question, incidentally. – Steven Stadnicki Dec 24 '11 at 15:07
up vote 1 down vote accepted

By definition, $n$ is the sum of the factorials of its digits. Since each digit of $n$ is at most 9, this can be at most $9!\cdot d$, where $d$ is the number of digits of $n$:

If $n$ is a factorion: $$ n=d_1d_2\cdots d_d\quad\Rightarrow\quad n= d_1!+\,d_2!\,+\cdots+ \,d_d!\le \underbrace{9!+\,9!+\,\cdots+\, 9!}_{d -\text{terms}}\le9!\cdot d $$

The lower bound is trivial, since $n$ has $d$ digits, it must be at least $10^{d-1}$.

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