Difference between revisions of "2020 AMC 12A Problems/Problem 10"
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==Solution 3 (Change of Base)== | ==Solution 3 (Change of Base)== | ||
Using the change of base formula on the RHS of the initial equation yields | Using the change of base formula on the RHS of the initial equation yields | ||
− | <cmath> \log_2{(\log_{16}{n})} = \frac{\log_2{(\log_4{n})}}{\log_2{4}} </cmath> | + | <cmath> \log_2{(\log_{16}{n})} = \frac{\log_2{(\log_4{n})}}{\log_2{4}}. </cmath> |
− | This means we can multiply each side by 2 for | + | This means we can multiply each side by <math>2</math> for |
− | <cmath> \log_2{(\log_{16}{n})^2} = \log_2{(\log_4{n})} </cmath> | + | <cmath> \log_2{(\log_{16}{n})^2} = \log_2{(\log_4{n})}. </cmath> |
Canceling out the logs gives | Canceling out the logs gives | ||
− | <cmath> (\log_{16}{n})^2=\log_4{n} </cmath> | + | <cmath> (\log_{16}{n})^2=\log_4{n}. </cmath> |
− | We use change of base on the RHS to see that <cmath> (\log_{16}{n})^2=\frac{ \log_{16}{n}}{\log_{16}{4}} | + | We use change of base on the RHS to see that |
+ | <cmath>\begin{align*} | ||
+ | (\log_{16}{n})^2&=\frac{ \log_{16}{n}}{\log_{16}{4}} \\ | ||
+ | (\log_{16}{n})^2&=2 \log_{16}{n}. | ||
+ | \end{align*}</cmath> | ||
+ | Substituting in <math> m = \log_{16}{n} </math> gives <math> m^2=2m, </math> so <math> m </math> is either <math>0</math> or <math>2.</math> Since <math> m=0 </math> yields no solution for <math>n</math> (since this would lead to use taking the log of <math>0</math>), we get <math> 2 = \log_{16}{n}, </math> or <math> n=16^2=256, </math> for the digit-sum of <math>2 + 5 + 6 = \boxed{\textbf{(E) }13}.</math> | ||
+ | |||
+ | ~aop2014 | ||
==Solution 4== | ==Solution 4== |
Revision as of 08:25, 15 August 2021
Contents
Problem
There is a unique positive integer such that
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, the original equation becomes
Using log property of addition, we can expand the parentheses into
Expanding the RHS and simplifying the logs without variables, we have
Subtracting
from both sides and adding
to both sides gives us
Multiplying by
exponentiating, and simplifying gives us
Adding the digits together, we have
~quacker88 (Solution)
~MRENTHUSIASM (Reformatting)
Solution 2
We know that, as the answer is an integer, must be some power of
Testing
yields
which does not work. We then try
giving us
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)
~MRENTHUSIASM (Reformatting)
Solution 3 (Change of Base)
Using the change of base formula on the RHS of the initial equation yields
This means we can multiply each side by
for
Canceling out the logs gives
We use change of base on the RHS to see that
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 the digit-sum of
~aop2014
Solution 4
Suppose Similarly, we have
Thus, we have
and
so
Plugging this in to either one of the expressions for
gives
, and the requested answer is
Solution 5
We will apply the following logarithmic identity:
which can be proven by the Change of Base Formula:
Note that
so we rewrite the original equation as follows:
from which
The sum of its digits is
~MRENTHUSIASM
Video Solution 1
~IceMatrix
Video Solution 2
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.