Does there exist a natural number $n$ such that $n!$ ends with exactly $2021$ zeros?
Problem
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Tags: number theory
rafaello
27.02.2021 22:11
We must have $$f(n)=\lfloor \frac{n}{5}\rfloor+\lfloor \frac{n}{25}\rfloor+\lfloor \frac{n}{125}\rfloor+\ldots=2021$$But if $n=8099$, then $f(8099)=1619+323+64+12+2=2020$. if $n=8100$, then $f(8100)=1620+324+64+12+2=2022$. And since $f(n)$ is non-decreasing on $n\in\mathbb N$, we see that there are $\boxed{\text{no}}$ such $n$.
MrOreoJuice
28.02.2021 07:53
$$\lfloor \frac{n}{5}\rfloor+\lfloor \frac{n}{25}\rfloor+\lfloor \frac{n}{125}\rfloor+ \cdots < \frac{n}{5} + \frac{n}{25} + \cdots$$$$ = \frac{\frac{n}{5}}{1 - \frac{1}{5}} = \frac{n}{4}$$$$\implies 2021 < \frac{n}{4}$$$$\implies n > 8084$$Now check numbers "near" 8084 to get an idea when there will be 2021 zeroes