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Question: What are the last $4$ nonzero digits of $2025!$ in base fourteen?

Note: This is not a contest/homework question.

I know how to find the number of trailing zeroes in base fourteen, so $2025!$ has:

$\lfloor{\frac{2025}{7}}\rfloor+\lfloor{\frac{2025}{7^{2}}}\rfloor+\lfloor{\frac{2025}{7^{3}}}\rfloor=289+41+5=335$ trailing zeroes.

But I was stucked here: Find the last $4$ nonzero digits of $2025!$ in base fourteen.

My instinct is to exclude all multiples of $7$ below or equal to $2025$ based on this answer, so all I have to do is to compute $(1\times2\times3\times4\times5\times6)(8\times9\times10\times11\times12\times13)\cdots(2017\times2018\times2019\times2020\times2021\times2022)(2024\times2025)\pmod{2401}$ to find the last $4$ nonzero digits of $2025!$ in base fourteen, as $2025!$ is certainly divisible by $16$, but I don't know how to continue.

Thirdy Yabata
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  • Does the answer here help? – Robert Israel May 10 '24 at 17:14
  • You need to compute the congruency class, $~\pmod{14^4}~$ of $~\dfrac{2025!}{14^{335}}.~$ Personally, I'm unsure how to do this. My approach, which might well fail, would be to first compute the congruency class $~\pmod{14},~$ and then compute the congruency class $~\pmod{14^2},~$ and so forth. – user2661923 May 10 '24 at 18:50

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