2020 AMC 12A Problem 10 — Solution | LIVE by Po-Shen Loh

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2020 AMC 12A Problem 10

Below is the professionally curated solution for Problem 10 of the 2020 AMC 12A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2020 AMC 12A solutions, or check the answer key.

All of the real AMC 8, AMC 10, AMC 12, and AIME problems in our complete solution collection are used with official legal permission of the Mathematical Association of America (MAA).

Concepts:logarithm substitution

Difficulty rating: 1500

10.

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 $n?$

$4$

$7$

$8$

$11$

$13$

Solution:

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.

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