Starting digits of 2^n

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: en.wikipedia.org/wiki/Poincar%C3%A9_recurrence_theorem –  WWright Dec 5 '10 at 21:01

1 Answer

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): en.wikipedia.org/wiki/Dirichlet's_approximation_theorem –  Qiaochu Yuan Dec 6 '10 at 16:35