Difference between revisions of "2021 AMC 12A Problems/Problem 18"
(→Video Solution by Punxsutawney Phil) |
(→Solution 1) |
||
Line 5: | Line 5: | ||
==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>. |
+ | |||
+ | -Lemonie | ||
+ | |||
==Video Solution by Punxsutawney Phil== | ==Video Solution by Punxsutawney Phil== | ||
https://youtu.be/8gGcj95rlWY | https://youtu.be/8gGcj95rlWY |
Revision as of 14:20, 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 .
-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.