AMC12 2020 A

AMC12 2020 A · Q10

AMC12 2020 A · Q10. It mainly tests Logarithms (rare), Manipulating equations.

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 $n$?

存在唯一的正整数 $n$ 使得 $\log_2(\log_{16} n) = \log_4(\log_4 n)$。 $n$ 的各位数字之和是多少？

(A) 4 4

(B) 7 7

(D) 11 11

(E) 13 13

Answer

Correct choice: (E)

正确答案：（E）

Solution

Converting all the logarithms to base 2 gives $$\log_2\left(\frac{1}{4} \log_2 n\right) = \frac{1}{2} \log_2\left(\frac{1}{2} \log_2 n\right).$$ Then $$\log_2\left(\frac{1}{4}\right) + \log_2(\log_2 n) = \frac{1}{2} \log_2\left(\frac{1}{2}\right) + \frac{1}{2} \log_2(\log_2 n).$$ This simplifies to $\log_2(\log_2 n) = 3$, so $n = 2^{(2^3)} = 256$. The requested sum of digits is $2 + 5 + 6 = 13$. OR Let the values of $\log_2(\log_{16} n)$ and $\log_4(\log_4 n)$ both be equal to the real number $x$. Then $2^x = \log_{16} n$ and hence $$n = 16^{(2^x)} = (4^2)^{(2^x)} = 4^{(2 \cdot 2^x)}.$$ Similarly, $4^x = \log_4 n$ and hence $n = 4^{(4^x)} = 4^{(2^x \cdot 2^x)}$. It must therefore be the case that $2 \cdot 2^x = 2^x \cdot 2^x$ and hence $x = 1$. Therefore $n = 4^4 = 256$, and the requested sum is $2 + 5 + 6 = 13$.

将所有对数转换为底数 2 有 $$\log_2\left(\frac{1}{4} \log_2 n\right) = \frac{1}{2} \log_2\left(\frac{1}{2} \log_2 n\right)。$$ 然后 $$\log_2\left(\frac{1}{4}\right) + \log_2(\log_2 n) = \frac{1}{2} \log_2\left(\frac{1}{2}\right) + \frac{1}{2} \log_2(\log_2 n)。$$ 这简化为 $\log_2(\log_2 n) = 3$，所以 $n = 2^{(2^3)} = 256$。要求的各位数字之和是 $2 + 5 + 6 = 13$。或者令 $\log_2(\log_{16} n)$ 和 $\log_4(\log_4 n)$ 的值都等于实数 $x$。则 $2^x = \log_{16} n$，从而 $$n = 16^{(2^x)} = (4^2)^{(2^x)} = 4^{(2 \cdot 2^x)}$$。类似地，$4^x = \log_4 n$，从而 $n = 4^{(4^x)} = 4^{(2^x \cdot 2^x)}$。因此必须有 $2 \cdot 2^x = 2^x \cdot 2^x$，故 $x = 1$。因此 $n = 4^4 = 256$，要求的和是 $2 + 5 + 6 = 13$。

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