Does there exists $n \in \mathbb{N}$ such that $3^n$ begins with $2019$?
I couldn't start the problem. If it ends with 2019 then I could have done using modulo $10000$ but how should I do this?
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1Note the suggested duplicate's solution can easily be adapted for powers of $3$ instead of $2$, and for $2019$ instead of $2013$. Also, another fairly closely related previous post is For every binary number x, is there a power of 3 beginning in x?. – John Omielan Feb 19 '20 at 06:31
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A number $N$ starts with $2019$ means $$2019\cdot 10^k\le N < 2020\cdot 10^k$$ for some positive integer $k$. Now assume $N=3^n$. We get $$2019 \cdot 10^k \le 3^n < 2020 \cdot 10^k$$ Taking $\log_{10}$ we get $$\log_{10}2019+ k \le n \log_{10}3< \log_{10}2020+ k $$ or $$\log_{10}2.019 +(k+3)\le n \log_{10}3 < \log_{10}2.02 +(k+3)$$ Now $\log_{10}2.019= 0.3501\ldots$ and $\log_{10}2.02= 0.3503\ldots$ So we want an $n$ so that the fractional part of $n\log_{10}3$ is between $\log_{10}2.019$ and $\log_{10}2.02$. There are infinitely many such $n$ (this is a general result, because $\log_{10}3$ is irrational). The first such $n$ is $686$, obtained by using a table of values. Check: $$3^{686}=2019206509720585277431746271306154908670993819273518833142815697810757237541803723362829096672564746373419373456697219770586393981431257564929415334936849526405125004207169538836514096413800834022649848544273677879327434359994550766440010778015818466039197731490675919383555118603990950770743845838678932637441771577577717979129$$
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