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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_theorem2010-12-05
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    Some special (or general) cases of this question: [2011](https://math.stackexchange.com/questions/46100/fractional-part-of-b-log-a), [2013](https://math.stackexchange.com/questions/544214/is-2k-2013-for-some-k), [7](https://math.stackexchange.com/questions/2230226/show-that-there-are-infinitely-many-powers-of-two-starting-with-the-digit-7). (At the [2011 question](https://math.stackexchange.com/questions/46100/fractional-part-of-b-log-a) I've left an answer with a constructive method.)2017-04-18

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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).

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    You 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_theorem2010-12-06