How many positive integers $n$ satisfy the following condition:
$(130n)^{50} > n^{100} > 2^{200}\ ?$
$\textbf{(A) } 0\qquad \textbf{(B) } 7\qquad \textbf{(C) } 12\qquad \textbf{(D) } 65\qquad \textbf{(E) } 125$
We're given $(130n)^{50} > n^{100} > 2^{200}$, so
$\sqrt[50]{(130n)^{50}} > \sqrt[50]{n^{100}} > \sqrt[50]{2^{200}}$ (because all terms are positive) and thus
$130n > n^2 > 2^4$
$130n > n^2 > 16$
Solving each part separately: 
$n^2 > 16 \Longrightarrow n > 4$
$130n > n^2 \Longrightarrow 130 > n$
So $4 < n < 130$. 
Therefore the answer is the number of [positive integers](https://artofproblemsolving.com/wiki/index.php/Positive_integer) over the interval $(4,130)$ which is $129-5+1 = \boxed{\textbf{(E) }125}$