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Does there exist a natural number $n$ such that the first four digits in the decimal representation of $3^n$ is $2019$?

I know to estimate the last few digits of any exponent by reducing modulo $10^k$, but how to obtain the first few digits? Note that computational tools are discouraged. Instead pure analysis must be used to arrive at the conclusion. Any hints? Thanks beforehand.

vidyarthi
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    Possibly a hint: More generally, we can ask about the density of the image of $n \mapsto {n\cdot \log_{10}(3)}$ in $[0,1)$. (Here, ${x}$ denotes the fractional part of $x$) – Brian Moehring Feb 05 '20 at 21:49
  • @BrianMoehring sorry, what do you imply by 'density' – vidyarthi Feb 05 '20 at 21:56
  • I think Brian means to ask if there's a value between any two given values – J. W. Tanner Feb 05 '20 at 21:58
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    In short, there is an open set $(a,b) \subset [0,1)$ such that $a < {n\cdot \log_{10}(3)} < b$ for some $n$ implies $3^n$ begins with whatever string of digits you want, including $2019.$ There may be another tool to study the first however many digits, but the fractional part of the logarithm is the only one I know of. – Brian Moehring Feb 05 '20 at 22:00
  • @BrianMoehring so the answer is in the affirmative right? but proving the density is hard, I suppose, I think it uses the rational approximation to the irrational $\log_{10}3$ right? – vidyarthi Feb 05 '20 at 22:01
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    The point is that $\log_{10}(3)$ is irrational, and the sequence of fractional parts of multiples of an irrational number is dense in $[0,1]$. – Robert Israel Feb 05 '20 at 22:04
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    @RobertIsrael so this implies if $\log_{10}a$ is irrational, then $a^n$ could start with any string of integers, right? – vidyarthi Feb 05 '20 at 22:06
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    Yes, it does. BTW, in this case $n=686$ works. – Robert Israel Feb 05 '20 at 22:07
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    In fact, more is known: the sequence of fractional parts of multiples of an irrational number is uniformly distributed in $[0,1]$, in the sense that the number of $n \in [0,N]$ where the fractional part is in $[a,b] \subseteq [0,1]$ is asymptotic to $(b-a)N$. See equidistribution theorem – Robert Israel Feb 05 '20 at 22:16

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