2014 AMC 12A Problem 18

Working from the outside, $f$ is defined exactly when $\log_4\!\left(\log_{\frac{1}{4}}\!\left(\log_{16}\!\left(\log_{\frac{1}{16}}x\right)\right)\right)$ $\gt0,$ which is equivalent to $\log_{\frac{1}{4}}\!\left(\log_{16}\!\left(\log_{\frac{1}{16}}x\right)\right)\gt1.$ Since the base $\tfrac14\lt1,$ this means $0\lt\log_{16}\!\left(\log_{\frac{1}{16}}x\right)\lt\tfrac14,$ hence $1\lt\log_{\frac{1}{16}}x\lt16^{\frac{1}{4}}=2.$ As $\tfrac{1}{16}\lt1,$ this reverses to $\left(\tfrac{1}{16}\right)^2\lt x\lt\left(\tfrac{1}{16}\right)^1,$ i.e. $\tfrac{1}{256}\lt x\lt\tfrac{1}{16}.$ The length is $\tfrac{1}{16}-\tfrac{1}{256}=\tfrac{15}{256},$ so $m+n=15+256=271.$ Thus, the correct answer is C.