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So I saw this problem in an Olympiad book, "Prove that for any natural number $n$, there exists a power of $2$ which contains $n$ in it. "

For example, $n=19$ is in $2^{13}=8192$, $n=24$ is in $2^{10}=1024$.

I tried solving it by Pigeonhole principle, but haven't made any progress. Any ideas?

Sudheesh Surendranath
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    I'd look for a power of two that starts with the digits of $n$. – Angina Seng Aug 19 '18 at 11:53
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    This should be a duplicate. Your claim follows from the fact, that the fractional part of $k\cdot log_{10}(2)$ is equidistributed in the interval $[0,1]$ and therefore $2^k$ can begin with every finite digit string (that does not start with $0$) – Peter Aug 19 '18 at 12:11

1 Answers1

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Here's a quick sketch of a proof:

  1. Prove $\log_{10}(2)$ is irrational.

  2. Prove that the fractional part of $n\alpha$, where $n \in \mathbb{N}$ and $\alpha$ is irrational, is dense in $(0,1)$

  3. Note that the fractional part of $\log(x)$ determines the first few digits, and then use the density of the fractional part of $n\alpha$ to prove that the first few digits of the number can be any $m \in \mathbb{N}$.

If you need more help, just ask!

Isaac Browne
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