The domain of the function $f(x)=\log_{\frac12}(\log_4(\log_{\frac14}(\log_{16}(\log_{\frac1{16}}x))))$ is an interval of length $\tfrac mn$, where $m$ and $n$ are relatively prime positive integers.  What is $m+n$?
$\textbf{(A) }19\qquad \textbf{(B) }31\qquad \textbf{(C) }271\qquad \textbf{(D) }319\qquad \textbf{(E) }511\qquad$
For all real numbers $a,b,$ and $c$ such that $b>0$ and $b\neq1,$ note that:


$\log_b a$ is defined if and only if $a>0.$
For $0b^c.$
$\log_b a>c$ if and only if $01,$ we conclude that:


$\log_b ac$ if and only if $a>b^c.$


Therefore, we have
\begin{align*} \log_{\frac12}(\log_4(\log_{\frac14}(\log_{16}(\log_{\frac1{16}}x)))) \text{ is defined} &\implies \log_4(\log_{\frac14}(\log_{16}(\log_{\frac1{16}}x)))>0 \\ &\implies \log_{\frac14}(\log_{16}(\log_{\frac1{16}}x))>1 \\ &\implies 0<\log_{16}(\log_{\frac1{16}}x)<\frac14 \\ &\implies 1<\log_{\frac1{16}}x<2 \\ &\implies \frac{1}{256}<x<\frac{1}{16}. \end{align*}
The domain of $f(x)$ is an interval of length $\frac{1}{16}-\frac{1}{256}=\frac{15}{256},$ from which the answer is $15+256=\boxed{\textbf{(C) }271}.$
Remark
This problem is quite similar to [2004 AMC 12A Problem 16](https://artofproblemsolving.com/wiki/index.php/2004_AMC_12A_Problems/Problem_16).
~MRENTHUSIASM