2020 AMC 12A Problem 10

Since $\log_{16} n = \tfrac12 \log_4 n,$ set $y = \log_4 n.$ The equation becomes $\log_2\left(\tfrac{y}{2}\right) = \log_4 y = \tfrac12 \log_2 y.$ Multiplying by $2$ gives $\log_2\left(\tfrac{y}{2}\right)^2 = \log_2 y,$ so $\left(\tfrac{y}{2}\right)^2 = y,$ which yields $y = 4.$ Then $\log_4 n = 4,$ so $n = 4^4 = 256,$ and the digit sum is $2 + 5 + 6 = 13.$ Thus, E is the correct answer.