Difference between revisions of "2020 AMC 12A Problems/Problem 10"
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<math>\textbf{(A) } 4 \qquad \textbf{(B) } 7 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 11 \qquad \textbf{(E) } 13</math> | <math>\textbf{(A) } 4 \qquad \textbf{(B) } 7 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 11 \qquad \textbf{(E) } 13</math> | ||
| − | ==Solution== | + | ==Solution 1== |
Any logarithm in the form <math>\log_{a^b} c = \frac{1}{b} \log_a c</math>. (this can be proved easily by using change of base formula to base <math>a</math>). | Any logarithm in the form <math>\log_{a^b} c = \frac{1}{b} \log_a c</math>. (this can be proved easily by using change of base formula to base <math>a</math>). | ||
Revision as of 15:27, 15 February 2021
Contents
Problem
There is a unique positive integer 
such that![\[\log_2{(\log_{16}{n})} = \log_4{(\log_4{n})}.\]](http://latex.artofproblemsolving.com/d/2/3/d23cbff85ac447661855a4f0344f9c162ce14b2e.png)
What is the sum of the digits of 
Solution 1
Any logarithm in the form 
. (this can be proved easily by using change of base formula to base 
).
so ![\[\log_2{(\log_{2^4}{n})} = \log_{2^2}{(\log_{2^2}{n})}\]](http://latex.artofproblemsolving.com/6/b/c/6bcb5e9212f349f1435c4621656e0b9d6a378acd.png)
becomes
![\[\log_2({\frac{1}{4}\log_{2}{n}}) = \frac{1}{2}\log_2({\frac{1}{2}\log_2{n}})\]](http://latex.artofproblemsolving.com/7/a/a/7aa4e9f55b54bb45db9b85e3d507b33d2f5ab5d6.png)
Using 
property of addition, we can expand the parentheses into
![\[\log_2{(\frac{1}{4})}+\log_2{(\log_{2}{n}}) = \frac{1}{2}(\log_2{(\frac{1}{2})} +\log_{2}{(\log_2{n})})\]](http://latex.artofproblemsolving.com/d/4/2/d420413d862673d0086ee45c71a767445a4b4218.png)
Expanding the RHS and simplifying the logs without variables, we have
![\[-2+\log_2{(\log_{2}{n}}) = -\frac{1}{2}+ \frac{1}{2}(\log_{2}{(\log_2{n})})\]](http://latex.artofproblemsolving.com/b/0/4/b0493c284840502aa92807d9d246c61f0f2953e2.png)
Subtracting 
from both sides and adding 
to both sides gives us
![\[\frac{1}{2}(\log_{2}{(\log_2{n})}) = \frac{3}{2}\]](http://latex.artofproblemsolving.com/8/a/7/8a7e05fe9f9869dfc11c714d49a9c7e8528a2312.png)
Multiplying by , raising the logs to exponents of base to get rid of the logs and simplifying gives us
![\[(\log_{2}{(\log_2{n})}) = 3\]](http://latex.artofproblemsolving.com/9/1/4/9146e6705d2138989b0e8ea9244e0f0e50d06521.png)
![\[2^{\log_{2}{(\log_2{n})}} = 2^3\]](http://latex.artofproblemsolving.com/b/d/7/bd7c2564ecf0eff4c92b32ef90fdcc3f37d16744.png)
![\[\log_2{n}=8\]](http://latex.artofproblemsolving.com/e/c/7/ec7803d033fc0d1ce3c2b76c81898695de7f2ddb.png)
![\[2^{\log_2{n}}=2^8\]](http://latex.artofproblemsolving.com/7/a/4/7a44490f9adaf3d685f5fd7c68e5828820d0addc.png)
![\[n=256\]](http://latex.artofproblemsolving.com/b/c/9/bc9693dd853b120ae972644d828454e9af81b9a5.png)
Adding the digits together, we have 
~quacker88
Solution 2
We know that, as the answer is an integer, must be some power of 
. Testing yields
![\[\log_2{(\log_{16}{16})} = \log_4{(\log_4{16})}\]](http://latex.artofproblemsolving.com/b/1/e/b1e43f1c1b129ffeab2722ae88c8caa3e114b3e1.png)
![\[\log_2{1} = \log_4{2}\]](http://latex.artofproblemsolving.com/5/a/f/5af17cab451a366a1e063a0e2c21d2bcc43f17c0.png)
![\[0 = \frac{1}{2}\]](http://latex.artofproblemsolving.com/6/b/5/6b58fd802199c2d48b69e6cd538c19563c731eb3.png)
which does not work. We then try 
, giving us
![\[\log_2{(\log_{16}{256})} = \log_4{(\log_4{256})}\]](http://latex.artofproblemsolving.com/8/8/f/88f35ad9bdd6833b7eeafa52ca0c70997400b67b.png)
![\[\log_2{2} = \log_4{4}\]](http://latex.artofproblemsolving.com/6/f/5/6f5cda2a51070dfd10e23aa9b999b6efcd6f54a2.png)
![\[1 = 1\]](http://latex.artofproblemsolving.com/a/d/1/ad1d4baf643b046ba23fdea73bd02ca2aa256ffb.png)
which holds true. Thus, 
, so the answer is 
.
(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
![\[\log_2{(\log_{16}{n})} = \frac{\log_2{(\log_4{n})}}{\log_2{4}}\]](http://latex.artofproblemsolving.com/c/c/5/cc586a7694cc2a544437ec5d10518d9df8c3dd3d.png)
This means we can multiply each side by 2 for
![\[\log_2{(\log_{16}{n})^2} = \log_2{(\log_4{n})}\]](http://latex.artofproblemsolving.com/2/a/9/2a9f201f3fbee3347411100e24b86ffcd7803b0d.png)
Canceling out the logs gives
![\[(\log_{16}{n})^2=\log_4{n}\]](http://latex.artofproblemsolving.com/f/0/6/f06ba75291cfeb01f39f0683cf569e42b26f7f6d.png)
We use change of base on the RHS to see that ![\[(\log_{16}{n})^2=\frac{ \log_{16}{n}}{\log_{16}{4}}\]](http://latex.artofproblemsolving.com/2/7/c/27c55fc88cccab45fd596e8b5d07bde1e6f63241.png)
or ![\[(\log_{16}{n})^2= 2 \log_{16}{n}\]](http://latex.artofproblemsolving.com/e/9/3/e93ca487654f5a1900ef682cf218415c31f446fe.png)
Substituting in 
gives 
, so 
is either 
or . Since 
yields no solution for (since this would lead to use taking the log of ), we get 
, or 
, for a sum of . ~aop2014
Solution 4
Suppose 
Similarly, we have 
Thus, we have ![\[16^{2^k}=(4^2)^{2^k}=4^{2^{k+1}}\]](http://latex.artofproblemsolving.com/1/1/a/11a51d99e1a0645dacaa922d362c286f8f99f902.png)
and ![\[4^{4^k}=4^{2^{2k}},\]](http://latex.artofproblemsolving.com/9/3/2/932234b759d70e8ebeaac843237349f46b8460b7.png)
so 
Plugging this in to either one of the expressions for gives , and the requested answer is 
Video Solution
~IceMatrix
Video Solution
https://youtu.be/RdIIEhsbZKw?t=814
~ pi_is_3.14
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. 
