How to prove there is a natural number $n$, the first four digits of $n! = 2018$

Question

I have a question, that is how to prove there is a natural number $n$, the first four digits of $n!$ is $2018$.

Though I found similar question of OEIS sequence A019799,but I still don't know how to prove.

Answer 1

If number $x$ has the form $$x = 2018\ldots,$$ then $$\log_{10} 2.018 \le\{\log_{10}{x}\} < \log_{10}2.019,$$ roughly $$0.3049211619< \{\log_{10}x\} < 0.3051363189;$$ where $\{z\}$ denotes fractional part of the number $z$.

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This way, we should prove that the sequence $$ S_n = \log_{10}(n!) = \sum_{k=1}^{n} \log_{10}k $$ contains element $S_{k_0}$, which fractional part belongs to the range $(0.3049211, \: 0.3051363)$.

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Direct calculation shows that such $k_0$-candidates are: $$670, \; 4906, \; 7083, \; 17787, \; \ldots$$ Therefore, the smallest $n!$ of the form $2018\ldots$ is $$670! = 2.0183491...\times 10^{1604};$$

WolframAlpha checking line:

 http://www.wolframalpha.com/input/?i=670!