Question

存在唯一的正整数 $n$ 满足 $\log_2(\log_{16} n) = \log_4(\log_4 n)\text{。}$

$n$ 的各位数字之和是多少？

There is a unique positive integer $n$ such that $\log_2(\log_{16} n) = \log_4(\log_4 n).$

What is the sum of the digits of $n?$

$4$

$7$

$8$

$11$

$13$

$\log_{16} n = \tfrac12 \log_4 n$ ，所以令 $y = \log_4 n$ 。

$\log_{16} n = \tfrac12 \log_4 n,$ so let $y = \log_4 n$

方程变为 $\log_2\left(\tfrac{y}{2}\right) = \tfrac12 \log_2 y$ ；平方来解出 $y$ 。

The equation becomes $\log_2\left(\tfrac{y}{2}\right) = \tfrac12 \log_2 y;$ square to solve for $y$

因为 $\log_{16} n = \tfrac12 \log_4 n$ ，令 $y = \log_4 n$ 。方程变为 $\log_2\left(\tfrac{y}{2}\right) = \log_4 y = \tfrac12 \log_2 y$ 。

两边乘以 $2$ 得到 $\log_2\left(\tfrac{y}{2}\right)^2 = \log_2 y$ ，所以 $\left(\tfrac{y}{2}\right)^2 = y$ ，从而 $y = 4$ 。

因此 $\log_4 n = 4$ ，所以 $n = 4^4 = 256$ ，各位数字之和为 $2 + 5 + 6 = 13$ 。

因此，正确答案是 E。

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.