A problem on pigeon hole principle The question is to show that 

there is a power of $2$ whose decimal representation starts with the digits $1999$. 

[Hint : apply pigeon hole principle]
How to approach this problem?
 A: I would use the equidistribution theorem, which says the multiples of an irrational number have their fractional parts equidistributed in the unit interval.  Given $n=2^k$ we have $\log_{10}n=k\log_{10}2$.  We know that $\log_{10}2$ is irrational.  We then just have to claim the existence of a $k$ such that the fractional part of $k\log_{10}2$ is between $\log_{10}1.999$ and $\log_{10}2$
A: Just to complete Ross Millikan's answer (+1 to him) we may compute the first terms of the continued fraction of $\log_{10} 2$ and get
$$ \log_{10}2 = [0;3, 3, 9, 2, 2, 4, \color{red}{6}, 2, 1, 1, 3, 1, 18,\ldots]$$
and by truncating the continued fraction at the highlighted term we get
$$ 13301\log_{10}2 = 4004-\varepsilon,\qquad 0<\varepsilon<3\cdot 10^{-5} $$
hence $\color{red}{2^{13302}}$ is a number with the wanted property.

Small addendedum: a proof of the irrationality of $\alpha=\log_{10}(2)$. Assuming $\alpha\in\mathbb{Q}$ we have $10^p=2^q$ for some $p,q\in\mathbb{N}^+$, but while $5\mid 10^p$, $5\nmid 2^q$, so we have a contradiction.
