, 2006

First we show there are arbitrarily long blocks of zeros in powers of $2$. To get at least $k$ zeros in $2^n$, this power must be of the form $y \cdot 10^{m+k} + z$ with $y, z$ positive integers and $z$ having at most $m$ digits, i.e. $z < 10^m$. Thus it is sufficient to find $n, m$ with $2^n$ having residue less than $10^m$ modulo $10^{m+k}$. By Euler's theorem for every positive integer $t$ we have (as $(2, 5^t) = 1$)
$$2^{\varphi(5^t)} \equiv 1 \pmod{5^t}.$$
Multiplying by $2^t$ we obtain
$$2^{t+\varphi(5^t)} \equiv 2^t \pmod{10^t}, \quad \text{thus} \quad 2^{t+\varphi(5^t)} \equiv y \cdot 10^t + 2^t$$
for some positive integer $y$. By previous we set $n = t + \varphi(5^t)$ and $m = t - k$. We would like to have $t$ with $2^t < 10^{t-k}$. Such a value of $t$ exists definitely, e.g. $t = 2k$ (since $2^{2k} = 4^k < 10^k$). It follows by presented, that in
$$2^{2k+\varphi(5^{2k})} = y \cdot 10^{2k} + 2^{2k}$$
there is a block of at least $k$ zeros.

Let us take (for a given $k$) power of $2$ (say $2^n$) containing a block of exactly $r$ zeros with $r \ge k$. We will study what happens to this block when considering subsequent powers, i.e. when multiplying the number with block by $2$. For some nonzero digits $a, b$ we have
$$2^n = \underbrace{\dots a \ 0 \ 0 \ \dots}_{y} \underbrace{\dots 0 \ b \ \dots}_{r \text{ zeros}} \underbrace{\dots}_{z} = y \cdot 10^{r+s} + z.$$
Thus $2^{n+1} = 2y \cdot 10^{r+s} + 2z$. The number $2z$ has either the same number of digits as $z$, or one more. Hence at the "right side" the block of zeros either does not cut down, or cut down by one zero. At the "left side" the block can only extend (when $y$ divisible by $5$). Globally the length of the block either decreases by $1$, or does not change, or increases. Similarly when we will multiply by $2$ again and again, the length of the block in every step decreases at most by $1$. Thus the only possibility how to avoid block of length $k$ is to remain the length more than $k$. This is anyway impossible. Namely $y$ has in its prime factorization the prime $5$ with some exponent, say $\alpha$. When we $\alpha$ times multiply $2^n$ by $2$, under the next multiplying the block will not extend at the "left side". At the "right side" at least under every fourth multiplying the block cut down (since $2^4 > 10$). Hence after sufficient number of steps we obtain the power of $2$ with a block of exactly $k$ zeros.