For any $n$ is there $n$ consecutive $0$ in the decimal expansion of $2^m$ for suitable $m \in \mathbb N$?

Question

We have define a function $f:\mathbb N \rightarrow \mathbb N$ such that $f(n)=$ { smallest $m \in \mathbb N$ such that decimal expansion of $2^m$ have $n$ consecutive $0$ }.
I computed some values of $f$:
$f(1)=10$ $\quad 2^{10}=1024$ have $1$ consecutive $0$.
$f(2)=53$ $\quad 2^{53}=9007199254740992$ have $2$ consecutive $0$...
$f(3)=242$
$f(4)=377$
$f(5)=1491$
$f(6)=1492$
$f(7)=6801$
$f(8)=14007$
.
.
.
Is it true that $f$ is a well defined function?!

Answer 1

Hint: Since $\log_{10}(2)$ is irrational, you can use Weyl's equidistribution or Dirichlet's approximation theorem to prove your result.