Difference between revisions of "2020 AMC 12A Problems/Problem 10"
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<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}} </cmath> or <cmath> (log_{16}{n})^2= 2 log_{16}{n}</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 a log cannot be equal to <math>0</math>), we get <math> 2 = log_{16}{n} </math>, or <math> n=16^2=256 </math>, for a sum of <math>2 + 5 + 6 = \boxed{\textbf{(E) }13}</math>. ~aop2014 | We use change of base on the RHS to see that <cmath> (log_{16}{n})^2=\frac{ log_{16}{n}}{\log_{16}{4}} </cmath> or <cmath> (log_{16}{n})^2= 2 log_{16}{n}</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 a log cannot be equal to <math>0</math>), we get <math> 2 = log_{16}{n} </math>, or <math> n=16^2=256 </math>, for a sum of <math>2 + 5 + 6 = \boxed{\textbf{(E) }13}</math>. ~aop2014 | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/fzZzGqNqW6U | ||
+ | |||
+ | ~IceMatrix | ||
==See Also== | ==See Also== |
Revision as of 16:32, 3 February 2020
Problem
There is a unique positive integer such that
What is the sum of the digits of
Solution
Any logarithm in the form . (this can be proved easily by using change of base formula to base
).
so
becomes
Using 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 , raising the logs to exponents of base
to get rid of the logs and simplifying gives us
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
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 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 2 for
Canceling out the logs gives
We use change of base on the RHS to see that
or
Substituting in
gives
, so
is either
or
. Since
yields no solution for
(since a log cannot be equal to
), we get
, or
, for a sum of
. ~aop2014
Video Solution
~IceMatrix
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.