2020 AMC 12A Problem 10

Attempt Problem 10 of the 2020 AMC 12A below, then check your answer against the professionally curated solution 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.

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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

\log_{16} n = \tfrac12 \log_4 n,

so let

y = \log_4 n

The equation becomes

\log_2\left(\tfrac{y}{2}\right) = \tfrac12 \log_2 y;

square to solve for

y

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,

\left(\tfrac{y}{2}\right)^2 = y,

which yields

y = 4.

Then

\log_4 n = 4,

n = 4^4 = 256,

and the digit sum is

2 + 5 + 6 = 13.

Thus, E is the correct answer.

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