mathmdmb wrote: Prove that $5^n$ has a block of $1976$ consecutive $0's$ in its decimal representation. $\log_{10}(5)$ is irrational and so $\{\{n\log_{10}(5)\},n\in\mathbb N\}$ is dense in $[0,1]$ So we can find $n$ such that $u=\{n\log_{10}5\}$ is as near as we want of $0$ and so $5^n=10^p10^u$ begins with $1$ and as many zeroes we want. Hence the result.

It has been posted that for a positive integer $a\geq 2$, such that $\log_{10} a$ is irrational, and some positive integer $N$, there exist infinitely many powers $a^n$ of $a$ whose decimal representations start with $N$. Since $\log_{10} 5$ is irrational, take $N = 10^{1976}$.

It is also possible to have a bunch of zeros close to the tail of decimal representation. See http://www.artofproblemsolving.com/Forum/viewtopic.php?f=56&t=358748 for details.

More powerful: For any prime $p$ we can found m such that decimal representation begins on arbitrary row of digits.

Sorry for question, but it this fact is true? For any irrational number $b$ and any $\epsilon > 0 $ we can found $m, n \in \mathbb{N} $ for which $0< \frac{m}{n}- b < \frac{\epsilon }{n}$. If this fact is true, then I have solution of my problem

But of course, dear Denis! See http://en.wikipedia.org/wiki/Dirichlet's_approximation_theorem. It proves that there exist infinitely many rationals $\dfrac {m} {n}$ such that $\left | \dfrac {m} {n} - b \right | < \dfrac {1} {n^2}$; all is left is to take $n > \dfrac {1} {\epsilon}$ so we can prolong the inequality to $\left | \dfrac {m} {n} - b \right | < \dfrac {1} {n^2} < \dfrac {\epsilon} {n}$. But have you read my reply (post #3)?

Yes, really, I don't see it, sorry