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Several of my students have noticed that the factorial of 23 is the first to contain at least one copy of each of the ten digits, and they wonder whether every sufficiently large factorial similarly contains at least one of each of the ten digits.

I would guess yes, but has this been proved?

Bernardo Recamán Santos
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  • $24!$ does not have any digit $5$ appearing. "Arbitrarily large factorial"... your question seems to be then... does there exist some $N$ such that for all $n\geq N$ you have $n!$ has all ten digits appearing in its base-10 representation... noting that $N$ would not be $23$. – JMoravitz Apr 05 '21 at 17:19
  • Clarified the issue. – Bernardo Recamán Santos Apr 05 '21 at 17:32
  • $25!$ has neither $6$ nor $7$ – saulspatz Apr 05 '21 at 17:35
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    $42!$ through $999!$ all have all $10$ digits, so perhaps $42$ is a good candidate. In python max(n for n in range(25,1000) if len(set(str(factorial(n)))) != 10) evaluates to $41$ – saulspatz Apr 05 '21 at 17:39
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    Though this doesn't answer the question: in case digits are uniformly distributed in the factorials, one notes that given sufficient digits (or given a large enough factorial), the probability of not finding all 10 digits can be made arbitrarily small – Rahul Madhavan Apr 05 '21 at 17:40
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    See https://math.stackexchange.com/questions/3613467/whats-the-next-base-ten-non-pandigital-factorial-number-after-41 – Barry Cipra Apr 05 '21 at 17:54
  • For $42\le n\le 3\cdot 10^4$ , $n!$ contains all the digits in the base $10$-representation, hence $42$ (the answer to everything :) ) should be the desired $N$. – Peter Apr 06 '21 at 07:27

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