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Is 40 the largest number for which all the 0 digits in the decimal form of $n!$ come at the end?

Motivation: My son considered learning all digits of 40! for my birthday. I told him that the best way to remember them would be to come up with a mnemonic. He asked me what word he should come up with if the digit is 0. The rest is history.

domotorp
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    Interesting question, but very difficult. I am not sure we even know if there are an infinite number of 0s in the decimal expansion of pi. (But it's believed this is the case) – Per Alexandersson May 28 '21 at 16:24
  • Is there an integer n which does not divide 10 such that for any k integer with 10 does not divide a = n * k and a has a zero in its decimal places ? – Dattier May 28 '21 at 16:39
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    And by the way, happy 40th birthday ;) – Loïc Teyssier May 28 '21 at 16:56
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    What follows doesn't answer your question, and probably won't contribute to an answer either, but a result by John Edward Maxfield -- A note on $N!$, Mathematics Magazine 43 #2 (March 1970), pp. 64-67 -- might be of interest: Given any positive integer $n,$ there exists a positive integer $N$ such that the decimal digits of $N!$ begin with all the decimal digits of $n,$ in their correct order. For more about this, see my answer to Short papers for undergraduate course on reading scholarly math. – Dave L Renfro May 28 '21 at 17:09
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    Using the heuristic that the digits of $n!$ are just random, followed by some zeroes at the end, you can calculate without too much trouble that the answer to your question is: probably yes. My (very crude) estimate puts it at a better than 99% chance. I don't know whether this heuristic is any good, and I doubt this way of thinking will lead to a real answer . . . but there you have it. – Will Brian May 28 '21 at 17:13
  • @Will, did your back-of-envelope calculation provide any estimate on when it would happen? – Jukka Kohonen May 28 '21 at 18:39
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    @JukkaKohonen: I mean that 40 probably is the largest number $n$ for which all the 0's in $n!$ appear at the end. – Will Brian May 28 '21 at 18:47
  • Oh, I understood it the wrong way! My bad. – Jukka Kohonen May 28 '21 at 18:51
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    @Loïc Thanks, though a bit early, and it is not the 40th, but the 815915283247897734345611269596115894272000000000th ;) – domotorp May 28 '21 at 19:07
  • I would learn a cyclic number instead, that he can use for magic tricks https://en.wikipedia.org/wiki/Cyclic_number – Pietro Majer May 28 '21 at 19:16
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    @domotorp: oh, that's what I feared... :D – Loïc Teyssier May 28 '21 at 19:44
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    One could call it the Zumkeller conjecture. – Spenser May 28 '21 at 20:14
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    Then I suppose this answers my question as 'known open problem'. – domotorp May 28 '21 at 20:48
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    Interesting... I could not find another example for $41 \leq n \leq 16000$. I wrote a small Python program for that purpose, which checked the $n$s in the previous range one by one. – Malkoun May 28 '21 at 21:03
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    As for the mnemonic, he can try to use 10 letters words, if they're not too uncommon in your native language. – Sylvain JULIEN May 29 '21 at 09:22
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    Ok, it seems you're Hungarian. Well, finding such words should not be too much of an issue :-) – Sylvain JULIEN May 29 '21 at 09:24
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    $41!$ seems to be the last factorial number with no digit 9 in it. It looks like for $n\ge 42$, the decimal representation of $n!$ will contain all ten digits, even after trailing zeros have been removed. Edit: Found OEIS entry. – Jeppe Stig Nielsen May 29 '21 at 16:32
  • @Malkoun I did a similar check before seeing your post, and ran it up to 40,000. OEIS A182049 mentioned by Jeppe Stig Nielsen implies it's been checked up to 100,000. I guess I could check somewhat further by letting a computer run through the weekend, but I doubt there's much point. – none May 29 '21 at 19:19
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    The title would be improved (and less puzzling) if it were replaced by the sentence that is the actual question. – A Stasinski May 31 '21 at 16:45
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    I changed the title to "$0$'s in 40!=...", that is, added "40!". This makes the title less cryptic within the list of questions. I regret you reverted it. – YCor May 31 '21 at 17:13
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    I prefer to keep an air of mystery about 815915283247897734345611269596115894272000000000. – domotorp May 31 '21 at 18:57
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    The point is that the question is not about the zeros in this number (which is a trivial question) but about the zeros in infinitely many other numbers. – A Stasinski Jun 01 '21 at 08:10
  • Maybe it will easy to solve in binary first. Where satisfied for 0!, 1!, 2!, 3! and 4!. – Pruthviraj Jun 11 '21 at 16:56

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