Difference between revisions of "2021 AMC 12A Problems/Problem 18"
MRENTHUSIASM (talk | contribs) (→Solution 4 (Most Comprehensive, Similar to Solution 3): Adjusted the row height a little.) |
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<math>\textbf{(A) }\frac{17}{32}\qquad\textbf{(B) }\frac{11}{16}\qquad\textbf{(C) }\frac{7}{9}\qquad\textbf{(D) }\frac{7}{6}\qquad\textbf{(E) }\frac{25}{11}\qquad</math> | <math>\textbf{(A) }\frac{17}{32}\qquad\textbf{(B) }\frac{11}{16}\qquad\textbf{(C) }\frac{7}{9}\qquad\textbf{(D) }\frac{7}{6}\qquad\textbf{(E) }\frac{25}{11}\qquad</math> | ||
− | ==Solution 1 | + | ==Solution 1== |
Looking through the solutions we can see that <math>f(\frac{25}{11})</math> can be expressed as <math>f(\frac{25}{11} \cdot 11) = f(11) + f(\frac{25}{11})</math> so using the prime numbers to piece together what we have we can get <math>10=11+f(\frac{25}{11})</math>, so <math>f(\frac{25}{11})=-1</math> or <math>\boxed{E}</math>. | Looking through the solutions we can see that <math>f(\frac{25}{11})</math> can be expressed as <math>f(\frac{25}{11} \cdot 11) = f(11) + f(\frac{25}{11})</math> so using the prime numbers to piece together what we have we can get <math>10=11+f(\frac{25}{11})</math>, so <math>f(\frac{25}{11})=-1</math> or <math>\boxed{E}</math>. | ||
Revision as of 12:41, 15 February 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
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 (Most Comprehensive, Similar to Solution 3)
We have the following important results:
for all positive integers
for all positive rational numbers
for all positive rational numbers
Proofs
Result can be shown by induction.
Result Since powers are just repeated multiplication, we will use result to prove result
Result For all positive rational numbers we have Therefore, we get So, result is true.
Result For all positive rational numbers we have It follows that and result is true.
For all positive integers and suppose and are their prime factorizations, respectively, we have
We apply function on each fraction in the choices:
Therefore, the answer is
~MRENTHUSIASM
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.