Difference between revisions of "2021 AMC 12A Problems/Problem 18"
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For all positive integers <math>x</math> and <math>y,</math> suppose <math>\prod_{k=1}^{m}p_k^{e_k}</math> and <math>\prod_{k=1}^{n}q_k^{d_k}</math> are their respective prime factorizations, we have | For all positive integers <math>x</math> and <math>y,</math> suppose <math>\prod_{k=1}^{m}p_k^{e_k}</math> and <math>\prod_{k=1}^{n}q_k^{d_k}</math> are their respective prime factorizations, we have | ||
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
− | f\left(\frac xy\right)&=f(x)+f\left(\frac 1y\right) & \text{by Result 1} \ | + | f\left(\frac xy\right)&=f(x)+f\left(\frac 1y\right) && \hspace{10mm}\text{by Result 1} \ |
− | &=f(x)-f(y) & \text{by Result 4} \ | + | &=f(x)-f(y) && \hspace{10mm}\text{by Result 4} \ |
&=f\left(\prod_{k=1}^{m}p_k^{e_k}\right)-f\left(\prod_{k=1}^{n}q_k^{d_k}\right) \ | &=f\left(\prod_{k=1}^{m}p_k^{e_k}\right)-f\left(\prod_{k=1}^{n}q_k^{d_k}\right) \ | ||
− | &=\left[\sum_{k=1}^{m}f\left(p_k^{e_k}\right)\right]-\left[\sum_{k=1}^{n}f\left(q_k^{d_k}\right)\right] & \text{by Result 1} \ | + | &=\left[\sum_{k=1}^{m}f\left(p_k^{e_k}\right)\right]-\left[\sum_{k=1}^{n}f\left(q_k^{d_k}\right)\right] && \hspace{10mm}\text{by Result 1} \ |
− | &=\left[\sum_{k=1}^{m}e_k f\left(p_k\right)\right]-\left[\sum_{k=1}^{n}d_k f\left(q_k\right)\right] &\hspace{10mm} \text{by Result 2} \ | + | &=\left[\sum_{k=1}^{m}e_k f\left(p_k\right)\right]-\left[\sum_{k=1}^{n}d_k f\left(q_k\right)\right] && \hspace{10mm}\text{by Result 2} \ |
&=\left[\sum_{k=1}^{m}e_k p_k \right]-\left[\sum_{k=1}^{n}d_k q_k \right]. | &=\left[\sum_{k=1}^{m}e_k p_k \right]-\left[\sum_{k=1}^{n}d_k q_k \right]. | ||
\end{align*}</cmath> | \end{align*}</cmath> |
Revision as of 08:45, 15 June 2021
- The following problem is from both the 2021 AMC 10A #18 and 2021 AMC 12A #18, so both problems redirect to this page.
Contents
[hide]- 1 Problem
- 2 Solution 1
- 3 Solution 2
- 4 Solution 3 (Deeper)
- 5 Solution 4 (Extremely Comprehensive, Similar to Solution 3)
- 6 Solution 5
- 7 Video Solution by Hawk Math
- 8 Video Solution by North America Math Contest Go Go Go Through Induction
- 9 Video Solution by Punxsutawney Phil
- 10 Video Solution by OmegaLearn (Using Functions and manipulations)
- 11 Video Solution by TheBeautyofMath
- 12 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 . Furthermore, suppose that also has the property that for every prime number . For which of the following numbers is ?
Solution 1
Looking through the solutions we can see that can be expressed as so using the prime numbers to piece together what we have we can get , so or .
-Lemonie
- awesomediabrine
Solution 2
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 (Deeper)
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 less 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 (Extremely Comprehensive, Similar to Solution 3)
Results
We have the following important results:
- for all positive rational numbers and positive integers
- for all positive rational numbers and positive integers
- for all positive rational numbers
~MRENTHUSIASM
Proofs
- Result 1: We can show Result 1 by induction.
- Result 2: Since positive powers are just repeated multiplication of the base, we will use Result 1 to prove Result 2:
- Result 3: For all positive rational numbers we have Therefore, we get from which Result 3 is true.
- Result 4: We have Therefore, we get from which Result 4 is true.
~MRENTHUSIASM
Solution
For all positive integers and suppose and are their respective prime factorizations, we have
We apply function to each fraction in the choices:
Therefore, the answer is
~MRENTHUSIASM
Solution 5
The problem gives us that f(p)=p. If we let a=p and b=1, we get f(p)=f(p)+f(1), which implies f(1)=0. Notice that the answer choices are all fractions, which means we will have to multiply an integer by a fraction to be able to solve it. Therefore, let's try plugging in fractions and try to solve them. Note that if we plug in a=p and b=1/p, we get f(1)=f(p)+f(1/p). We can solve for f(1/p) as -f(p)! This gives us the information we need to solve the problem. Testing out the answer choices gives us the answer of E.
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
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.