Difference between revisions of "2020 AMC 12A Problems/Problem 10"
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==Solution 2 (Properties of Logarithms)== | ==Solution 2 (Properties of Logarithms)== | ||
− | We will | + | We will apply the following logarithmic identity: |
<cmath>\log_{p^k}{q^k}=\log_{p}{q},</cmath> | <cmath>\log_{p^k}{q^k}=\log_{p}{q},</cmath> | ||
− | which can be proven by the [[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> | + | 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: | Note that <math>\log_{16}{n}\neq0,</math> so we rewrite the original equation as follows: | ||
<cmath>\begin{align*} | <cmath>\begin{align*} |
Revision as of 01:32, 30 January 2023
Contents
Problem
There is a unique positive integer such thatWhat 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.