2021 AMC 12A Problems/Problem 18
- The following problem is from both the 2021 AMC 10A #18 and 2021 AMC 12A #18, so both problems redirect to this page.
Contents
- 1 Problem
- 2 Solution 1 (Intuitive)
- 3 Solution 2 (Specific)
- 4 Solution 3 (Generalized)
- 5 Solution 4 (Generalized)
- 6 Solution 5 (Quick, Dirty, and Frantic Last Hope)
- 7 Solution 6 (Rushed Generalization)
- 8 Video Solution by Hawk Math
- 9 Video Solution by North America Math Contest Go Go Go Through Induction
- 10 Video Solution by Punxsutawney Phil
- 11 Video Solution by OmegaLearn (Using Functions and Manipulations)
- 12 Video Solution by TheBeautyofMath
- 13 Video Solution (Quick and Easy)
- 14 See also
Problem
Let be a function defined on the set of positive rational numbers with the property that for all positive rational numbers and . Suppose that also has the property that for every prime number . For which of the following numbers is ?
Solution 1 (Intuitive)
From the answer choices, note that On the other hand, we have Equating the expressions for produces from which Therefore, the answer is
Remark
Similarly, we can find the outputs of at the inputs of the other answer choices: Alternatively, refer to Solutions 2 and 4 for the full processes.
~Lemonie ~awesomediabrine ~MRENTHUSIASM
Solution 2 (Specific)
We know that . By transitive, we have Subtracting from both sides gives Also In we have .
In we have .
In we have .
In we have .
In we have .
Thus, our answer is .
~JHawk0224 ~awesomediabrine
Solution 3 (Generalized)
Consider the rational , for integers. We have . So . Let be a prime. Notice that . And . So if , . We simply need this to be greater than what we have for . Notice that for answer choices and , the numerator has fewer prime factors than the denominator, and so they are less likely to work. We check first, and it works, therefore the answer is .
~yofro
Solution 4 (Generalized)
We derive the following properties of
- By induction, we have for all positive rational numbers and positive integers
Since positive powers are just repeated multiplication of the base, it follows that for all positive rational numbers and positive integers
- For all positive rational numbers we have from which
- For all positive rational numbers we have from which
For all positive integers and suppose and are their respective prime factorizations. We get We apply to each fraction in the answer choices: Therefore, the answer is
~MRENTHUSIASM
Solution 5 (Quick, Dirty, and Frantic Last Hope)
Note that answer choices through are whereas is Because the functional equation is related to primes, we hope that the uniqueness of answer choice is enough.
~OliverA
Solution 6 (Rushed Generalization)
If f(a b) = f(a) + f(b), and if f(p) = p, then f(p p) = 2p. You can do this multiple times (Ex: f(p^3) = 3p). You can quickly assume then, that f(p^n) = np. Thus the answer choices can then be rewritten as the product of a prime and another prime to the negative power. Answer choices A-C are straightforward. For D, you can rewrite as . When you get to E, you get f(25) + f(), which is 10 - 11, which is -1. So the answer is
~Zeeshan12
Video Solution by Hawk Math
https://www.youtube.com/watch?v=dvlTA8Ncp58
Video Solution by North America Math Contest Go Go Go Through Induction
https://www.youtube.com/watch?v=ffX0fTgJN0w&list=PLexHyfQ8DMuKqltG3cHT7Di4jhVl6L4YJ&index=12
Video Solution by Punxsutawney Phil
Video Solution by OmegaLearn (Using Functions and Manipulations)
~ pi_is_3.14
Video Solution by TheBeautyofMath
~IceMatrix
Video Solution (Quick and Easy)
~Education, the Study of Everything
See also
2021 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
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 10 Problems and Solutions |
2021 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 17 |
Followed by Problem 19 |
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.