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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.

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    $\begingroup$ And by the way, happy 40th birthday ;) $\endgroup$
    – Loïc Teyssier
    Commented May 28, 2021 at 16:56
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    $\begingroup$ 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. $\endgroup$
    – Dave L Renfro
    Commented May 28, 2021 at 17:09
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    $\begingroup$ 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. $\endgroup$
    – Will Brian
    Commented May 28, 2021 at 17:13
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    $\begingroup$ One could call it the Zumkeller conjecture. $\endgroup$
    – Spenser
    Commented May 28, 2021 at 20:14
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    $\begingroup$ I prefer to keep an air of mystery about 815915283247897734345611269596115894272000000000. $\endgroup$
    – domotorp
    Commented May 31, 2021 at 18:57

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