Prove that for any finite sequence of decimal digits, there exists an $n$ such that the decimal expansion of $2^n$ begins with these digits.
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hmmm... you may want to look up Poincare's recurrence theorem: http://en.wikipedia.org/wiki/Poincar%C3%A9_recurrence_theorem – WWright Dec 05 '10 at 21:01
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Some special (or general) cases of this question: 2011, 2013, 7. (At the 2011 question I've left an answer with a constructive method.) – ShreevatsaR Apr 18 '17 at 22:35
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Take $\log_{10} (2^n) = n \log_{10} 2$, note that $\log_{10} 2$ is irrational, and use the equidistribution theorem (or prove what you want directly using the pigeonhole principle).
Yuval Filmus
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13You don't need equidistribution; this follows from Dirichlet's approximation theorem (which is the reason the pigeonhole principle is named after Dirichlet): http://en.wikipedia.org/wiki/Dirichlet's_approximation_theorem – Qiaochu Yuan Dec 06 '10 at 16:35