Difference between revisions of "2021 AMC 12A Problems/Problem 18"
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− | Consider the rational <math>\frac{a}{b}</math>, for <math>a,b</math> integers. We have <math>f(a)=f\left(\frac{a}{b}\cdot b\right)=f\left(\frac{a}{b}\right)+f(b)</math>. So <math>f\left(\frac{a}{b}\right)=f(a)-f(b)</math>. Let <math>p</math> be a prime. Notice that <math>f(p^k)=kf(p)</math>. And <math>f(p)= | + | Consider the rational <math>\frac{a}{b}</math>, for <math>a,b</math> integers. We have <math>f(a)=f\left(\frac{a}{b}\cdot b\right)=f\left(\frac{a}{b}\right)+f(b)</math>. So <math>f\left(\frac{a}{b}\right)=f(a)-f(b)</math>. Let <math>p</math> be a prime. Notice that <math>f(p^k)=kf(p)</math>. And <math>f(p)=p</math>. So if <math>a=p_1^{a_1}p_2^{a_2}\cdots p_k^{a_k}</math>, <math>f(a)=a_1p_1+a_2p_2+....+a_kp_k</math>. We simply need this to be greater than what we have for <math>f(b)</math>. Notice that for answer choices <math>A,B,C, </math> and <math>D</math>, the numerator <math>(a)</math> has less prime factors than the denominator, and so none of these work (upon checking). Therefore, the answer is <math>\boxed{\textbf{(E)}}</math>. |
~yofro | ~yofro |
Revision as of 14:02, 12 February 2021
Contents
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
Solution 2
We know that . Adding to both sides, we get Also In we have .
In we have .
In we have .
In we have .
In we have .
Thus, our answer is ~JHawk0224
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 none of these work (upon checking). Therefore, the answer is .
~yofro
Video Solution by Hawk Math
https://www.youtube.com/watch?v=dvlTA8Ncp58
Video Solution by Punxsutawney Phil
Video Solution by OmegaLearn (Using Functions and manipulations)
~ pi_is_3.14
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.