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Difference between revisions of "2020 AMC 12A Problems/Problem 10"

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<cmath>n=256</cmath>
 
<cmath>n=256</cmath>
  
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Adding the digits together, we have <math>2+5+6=\boxed{\textbf{E) }13}</math> ~quacker88
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Adding the digits together, we have <math>2+5+6=\boxed{\textbf{(E) }13}</math> ~quacker88
 +
 
 +
==Solution 2==
 +
We know that, as the answer is an integer, <math>n</math> must be some power of <math>16</math>. Testing <math>16</math> yields
 +
<cmath> \log_2{(\log_{16}{16})} = \log_4{(\log_4{16})} </cmath>
 +
<cmath> \log_2{1} = \log_4{2}</cmath>
 +
<cmath> 0 = \frac{1}{2}</cmath>
 +
which does not work. We then try <math>256</math>, giving us
 +
 
 +
<cmath> \log_2{(\log_{16}{256})} = \log_4{(\log_4{256})} </cmath>
 +
<cmath> \log_2{2} = \log_4{4}</cmath>
 +
<cmath> 1 = 1 </cmath>
 +
which holds true. Thus, <math>n = 256</math>, so the answer is <math>2 + 5 + 6 = \boxed{\textbf{(E) }13}</math>.
 +
 
 +
(Don't use this technique unless you absolutely need to! Guess and check methods aren't helpful for learning math.)
 +
 
 +
~ciceronii
  
 
==See Also==
 
==See Also==

Revision as of 14:45, 1 February 2020

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

See Also

2020 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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