Difference between revisions of "2021 AMC 12A Problems/Problem 18"
(→Solution 1) |
Sugar rush (talk | contribs) |
||
Line 1: | Line 1: | ||
==Problem== | ==Problem== | ||
− | Let <math>f</math> be a function defined on the set of positive rational numbers with the property that <math>f(a\cdot b) = f(a)+f(b)</math> for all positive rational numbers <math>a</math> and <math>b</math>. Furthermore, suppose that <math>f</math> also has the property that <math>f(p)=p</math> for every prime number <math>p</math>. For which of the following | + | Let <math>f</math> be a function defined on the set of positive rational numbers with the property that <math>f(a\cdot b) = f(a)+f(b)</math> for all positive rational numbers <math>a</math> and <math>b</math>. Furthermore, suppose that <math>f</math> also has the property that <math>f(p)=p</math> for every prime number <math>p</math>. For which of the following numbers <math>x</math> is <math>f(x) < 0</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> | <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> |
Revision as of 14:28, 11 February 2021
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
Video Solution by Punxsutawney Phil
See also
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.