Difference between revisions of "2021 AMC 12A Problems/Problem 18"
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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== | ||
| + | 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>. | ||
==Video Solution by Punxsutawney Phil== | ==Video Solution by Punxsutawney Phil== | ||
https://youtu.be/8gGcj95rlWY | https://youtu.be/8gGcj95rlWY | ||
Revision as of 15:20, 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 values 
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 
.
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. 
