2020 AMC 12A Problems/Problem 10

Problem

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 $n?$

$\textbf{(A) } 4 \qquad \textbf{(B) } 7 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 11 \qquad \textbf{(E) } 13$

Solution

Any logarithm in the form $\log_{a^b} c = \frac{1}{b} \log_a c$ . (this can be proved easily by using change of base formula to base $a$ ).

so $\log_2{(\log_{2^4}{n})} = \log_{2^2}{(\log_{2^2}{n})}$

becomes

$\log_2({\frac{1}{4}\log_{2}{n}}) = \frac{1}{2}\log_2({\frac{1}{2}\log_2{n}})$

Using $\log$ property of addition, we can expand the parentheses into

$\log_2{(\frac{1}{4})}+\log_2{(\log_{2}{n}}) = \frac{1}{2}(\log_2{(\frac{1}{2})} +\log_{2}{(\log_2{n})})$

Expanding the RHS and simplifying the logs without variables, we have

$-2+\log_2{(\log_{2}{n}}) = -\frac{1}{2}+ \frac{1}{2}(\log_{2}{(\log_2{n})})$

Subtracting $\frac{1}{2}(\log_{2}{(\log_2{n})})$ from both sides and adding $2$ to both sides gives us

$\frac{1}{2}(\log_{2}{(\log_2{n})}) = \frac{3}{2}$

Multiplying by $2$ , raising the logs to exponents of base $2$ to get rid of the logs and simplifying gives us

$(\log_{2}{(\log_2{n})}) = 3$

$2^{\log_{2}{(\log_2{n})}} = 2^3$

$\log_2{n}=8$

$2^{\log_2{n}}=2^8$

$n=256$

Adding the digits together, we have $2+5+6=\boxed{\textbf{(E) }13}$ ~quacker88

Solution 2

We know that, as the answer is an integer, $n$ must be some power of $16$ . Testing $16$ yields $\log_2{(\log_{16}{16})} = \log_4{(\log_4{16})}$ $\log_2{1} = \log_4{2}$ $0 = \frac{1}{2}$ which does not work. We then try $256$ , giving us

$\log_2{(\log_{16}{256})} = \log_4{(\log_4{256})}$ $\log_2{2} = \log_4{4}$ $1 = 1$ which holds true. Thus, $n = 256$ , so the answer is $2 + 5 + 6 = \boxed{\textbf{(E) }13}$ .

(Don't use this technique unless you absolutely need to! Guess and check methods aren't helpful for learning math.)

~ciceronii

Solution 3-Change of Base

Using the change of base formula on the RHS of the initial equation yields $\log_2{(\log_{16}{n})} = \frac{\log_2{(\log_4{n})}}{\log_2{4}}$ This means we can multiply each side by 2 for $\log_2{(\log_{16}{n})^2} = \log_2{(\log_4{n})}$ Canceling out the logs gives $(\log_{16}{n})^2=\log_4{n}$ We use change of base on the RHS to see that $(\log_{16}{n})^2=\frac{ \log_{16}{n}}{\log_{16}{4}}$ or $(\log_{16}{n})^2= 2 \log_{16}{n}$ Substituting in $m = \log_{16}{n}$ gives $m^2=2m$ , so $m$ is either $0$ or $2$ . Since $m=0$ yields no solution for $n$ (since a log cannot be equal to $0$ ), we get $2 = \log_{16}{n}$ , or $n=16^2=256$ , for a sum of $2 + 5 + 6 = \boxed{\textbf{(E) }13}$ . ~aop2014

Solution 4

Suppose $\log_2(\log_{16}n)=k\implies\log_{16}n=2^k\implies n=16^{2^k}.$ Similarly, we have $\log_4(\log_4 n)=k\implies \log_4 n=4^k\implies n=4^{4^k}.$ Thus, we have $16^{2^k}=(4^2)^{2^k}=4^{2^{k+1}}$ and $4^{4^k}=4^{2^{2k}},$ so $k+1=2k\implies k=1.$ Plugging this in to either one of the expressions for $n$ gives $256$ , and the requested answer is $2+5+6=\boxed{\textbf{(E) }13}.$

Video Solution

https://youtu.be/fzZzGqNqW6U

~IceMatrix

2020 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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