Difference between revisions of "2020 AMC 12A Problems/Problem 10"
(→Solution) |
MRENTHUSIASM (talk | contribs) (→Solution 2 (Properties of Logarithms)) |
||
(54 intermediate revisions by 13 users not shown) | |||
Line 1: | Line 1: | ||
− | == | + | ==Problem== |
− | + | There is a unique positive integer <math>n</math> such that<cmath>\log_2{(\log_{16}{n})} = \log_4{(\log_4{n})}.</cmath>What is the sum of the digits of <math>n?</math> | |
− | + | <math>\textbf{(A) } 4 \qquad \textbf{(B) } 7 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 11 \qquad \textbf{(E) } 13</math> | |
− | + | ==Solution 1 (Properties of Logarithms)== | |
− | <cmath>\log_2{\frac{1}{4}(\log_{2}{n})} = \frac{1}{2}\log_2{\frac{1}{2}(\log_2{n})}.</cmath> | + | We can use the fact that <math>\log_{a^b} c = \frac{1}{b} \log_a c.</math> This can be proved by using [[change of base formula]] to base <math>a.</math> |
+ | |||
+ | So, the original equation <math>\log_2{(\log_{2^4}{n})} = \log_{2^2}{(\log_{2^2}{n})}</math> becomes <cmath>\log_2\left({\frac{1}{4}\log_{2}{n}}\right) = \frac{1}{2}\log_2\left({\frac{1}{2}\log_2{n}}\right).</cmath> | ||
+ | Using log property of addition, we expand both sides and then simplify: | ||
+ | <cmath>\begin{align*} | ||
+ | \log_2{\frac{1}{4}}+\log_2{(\log_{2}{n}}) &= \frac{1}{2}\left[\log_2{\frac{1}{2}} +\log_{2}{(\log_2{n})}\right] \\ | ||
+ | \log_2{\frac{1}{4}}+\log_2{(\log_{2}{n}}) &= \frac{1}{2}\left[-1 +\log_{2}{(\log_2{n})}\right] \\ | ||
+ | -2+\log_2{(\log_{2}{n}}) &= -\frac{1}{2}+ \frac{1}{2}(\log_{2}{(\log_2{n})}). | ||
+ | \end{align*}</cmath> | ||
+ | Subtracting <math>\frac{1}{2}(\log_{2}{(\log_2{n})})</math> from both sides and adding <math>2</math> to both sides gives us <cmath>\frac{1}{2}(\log_{2}{(\log_2{n})}) = \frac{3}{2}.</cmath> | ||
+ | Multiplying by <math>2,</math> exponentiating, and simplifying gives us | ||
+ | <cmath>\begin{align*} | ||
+ | \log_{2}{(\log_2{n})} &= 3 \\ | ||
+ | \log_2{n}&=8 \\ | ||
+ | n&=256. | ||
+ | \end{align*}</cmath> | ||
+ | Adding the digits together, we have <math>2+5+6=\boxed{\textbf{(E) } 13}.</math> | ||
+ | |||
+ | ~quacker88 (Solution) | ||
+ | |||
+ | ~MRENTHUSIASM (Reformatting) | ||
+ | |||
+ | ==Solution 2 (Properties of Logarithms)== | ||
+ | We will apply the following logarithmic identity: | ||
+ | <cmath>\log_{p^k}{q^k}=\log_{p}{q},</cmath> | ||
+ | which can be proven by the [[change of base formula | Change of Base Formula]]: <cmath>\log_{p^k}{q^k}=\frac{\log_{p}{q^k}}{\log_{p}{p^k}}=\frac{k\log_{p}{q}}{k}=\log_{p}{q}.</cmath> | ||
+ | Note that <math>\log_{16}{n}\neq0,</math> so we rewrite the original equation as follows: | ||
+ | <cmath>\begin{align*} | ||
+ | \log_4{(\log_{16}{n})^2}&=\log_4{(\log_4{n})} \\ | ||
+ | (\log_{16}{n})^2&=\log_4{n} \\ | ||
+ | (\log_{16}{n})^2&=\log_{16}{n^2} \\ | ||
+ | (\log_{16}{n})^2&=2\log_{16}{n} \\ | ||
+ | \log_{16}{n}&=2, | ||
+ | \end{align*}</cmath> | ||
+ | from which <math>n=16^2=256.</math> The sum of its digits is <math>2+5+6=\boxed{\textbf{(E) } 13}.</math> | ||
+ | |||
+ | ~MRENTHUSIASM | ||
+ | |||
+ | ==Solution 3 (Properties of Logarithms)== | ||
+ | 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> | ||
+ | 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> | ||
+ | Canceling out the logs gives | ||
+ | <cmath> (\log_{16}{n})^2=\log_4{n}. </cmath> | ||
+ | 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 (Exponential Form)== | ||
+ | Suppose <math>\log_2(\log_{16}n)=k\implies\log_{16}n=2^k\implies n=16^{2^k}.</math> Similarly, we have <math>\log_4(\log_4 n)=k\implies \log_4 n=4^k\implies n=4^{4^k}.</math> Thus, we have <cmath>16^{2^k}=(4^2)^{2^k}=4^{2^{k+1}}</cmath> and <cmath>4^{4^k}=4^{2^{2k}},</cmath> so <math>k+1=2k\implies k=1.</math> Plugging this in to either one of the expressions for <math>n</math> gives <math>256</math>, and the requested answer is <math>2+5+6=\boxed{\textbf{(E) } 13}.</math> | ||
+ | |||
+ | ==Solution 5 (Guess and Check)== | ||
+ | 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>\begin{align*} | ||
+ | \log_2{(\log_{16}{16})} &= \log_4{(\log_4{16})} \\ | ||
+ | \log_2{1} &= \log_4{2} \\ | ||
+ | 0 &= \frac{1}{2}, | ||
+ | \end{align*}</cmath> | ||
+ | which does not work. We then try <math>256,</math> giving us | ||
+ | <cmath>\begin{align*} | ||
+ | \log_2{(\log_{16}{256})} &= \log_4{(\log_4{256})} \\ | ||
+ | \log_2{2} &= \log_4{4} \\ | ||
+ | 1 &= 1, | ||
+ | \end{align*}</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 (Solution) | ||
+ | |||
+ | ~MRENTHUSIASM (Reformatting) | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/fzZzGqNqW6U | ||
+ | |||
+ | ~IceMatrix | ||
+ | |||
+ | == Video Solution by OmegaLearn == | ||
+ | https://youtu.be/RdIIEhsbZKw?t=814 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/EnyzIHcJ8Aw | ||
+ | |||
+ | ~Education, the Study of Everything | ||
+ | |||
+ | ==See Also== | ||
+ | |||
+ | {{AMC12 box|year=2020|ab=A|num-b=9|num-a=11}} | ||
+ | {{MAA Notice}} |
Latest revision as of 02:32, 30 January 2023
Contents
Problem
There is a unique positive integer such that
What is the sum of the digits of
Solution 1 (Properties of Logarithms)
We can use the fact that This can be proved by using change of base formula to base
So, the original equation becomes
Using log property of addition, we expand both sides and then simplify:
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 (Properties of Logarithms)
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
Solution 3 (Properties of Logarithms)
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 (Exponential Form)
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 (Guess and Check)
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)
Video Solution
~IceMatrix
Video Solution by OmegaLearn
https://youtu.be/RdIIEhsbZKw?t=814
~ pi_is_3.14
Video Solution
~Education, the Study of Everything
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.