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 $0$...
$f(3)=242$
$f(4)=377$
$f(5)=1491$
$f(6)=1492$
$f(7)=6801$
$f(8)=14007$
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Is it true that $f$ is a well defined function?!
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1See this. – David Mitra Jul 17 '15 at 19:30
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Just a pretty obvious comment, but you could rephrase your question as to whether for each $n$, you can find a power $2^m$ that has $n$ or more consecutive 0's in the expansion of $2^m$. The minimality condition in your definition automatically follows. – user2566092 Jul 17 '15 at 19:32
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@DavidMitra You basically knocked the question out of the park. In my opinion you should post as an answer, with the quick reasoning to make the desired deduction here. – user2566092 Jul 17 '15 at 19:35
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@user2566092: It's possible $ { $smallest $m \in \mathbb N$ such that decimal expansion of $2^m$ have $n$ consecutive $0}= \phi$ ! – Jul 17 '15 at 19:36
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1@2000 The post has that obvious error. Regardless, it holds for any "decimal digits" with a non-zero first digit. Here is a related post that gives more detail for a specific case. – David Mitra Jul 17 '15 at 19:44
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@DavidMitra : O! Thanks! – Jul 17 '15 at 19:46
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Hint: Since $\log_{10}(2)$ is irrational, you can use Weyl's equidistribution or Dirichlet's approximation theorem to prove your result.
Eric Naslund
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