Difference between revisions of "2021 AMC 12A Problems/Problem 18"

(Video Solution by Punxsutawney Phil)
(Solution 1)
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==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>.
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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 $f$ be a function defined on the set of positive rational numbers with the property that $f(a\cdot b) = f(a)+f(b)$ for all positive rational numbers $a$ and $b$. Furthermore, suppose that $f$ also has the property that $f(p)=p$ for every prime number $p$. For which of the following values $x$ is $f(x) < 0$?

$\textbf{(A) }\frac{17}{32}\qquad\textbf{(B) }\frac{11}{16}\qquad\textbf{(C) }\frac{7}{9}\qquad\textbf{(D) }\frac{7}{6}\qquad\textbf{(E) }\frac{25}{11}\qquad$

Solution 1

Looking through the solutions we can see that $f(\frac{25}{11})$ can be expressed as $f(\frac{25}{11} \cdot 11) = f(11) + f(\frac{25}{11})$ so using the prime numbers to piece together what we have we can get $10=11+f(\frac{25}{11})$, so $f(\frac{25}{11})=-1$ or $\boxed{E}$.

-Lemonie

Video Solution by Punxsutawney Phil

https://youtu.be/8gGcj95rlWY

See also

2021 AMC 12A (ProblemsAnswer KeyResources)
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

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