2021 AMC 12A Problems/Problem 18

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 numbers $x$ 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

Solution 2

We know that $f(2)=2$ . Adding $f(1)$ to both sides, we get $\begin{align*} f(2)+f(1)&=2+f(1)\\ f(2)&=2+f(1)\\ 2&=2+f(1)\\ f(1)&=0 \end{align*}$ Also $f(2)+f\left(\frac{1}{2}\right)=f(1)=0 \implies 2+f\left(\frac{1}{2}\right)=0 \implies f\left(\frac{1}{2}\right) = -2$ $f(3)+f\left(\frac{1}{3}\right)=f(1)=0 \implies 3+f\left(\frac{1}{3}\right)=0 \implies f\left(\frac{1}{3}\right) = -3$ $f(11)+f\left(\frac{1}{11}\right)=f(1)=0 \implies 11+f\left(\frac{1}{11}\right)=0 \implies f\left(\frac{1}{11}\right) = -11$ In $\textbf{(A)}$ we have $f\left(\frac{17}{32}\right)=17+5f\left(\frac{1}{2}\right)=17-5(2)=7$ .

In $\textbf{(B)}$ we have $f\left(\frac{11}{16}\right)=11+4f\left(\frac{1}{2}\right)=11-4(2)=3$ .

In $\textbf{(C)}$ we have $f\left(\frac{7}{9}\right)=7+2f\left(\frac{1}{3}\right)=7-2(3)=1$ .

In $\textbf{(D)}$ we have $f\left(\frac{7}{6}\right)=7+f\left(\frac{1}{2}\right)+f\left(\frac{1}{3}\right)=7-2-3=2$ .

In $\textbf{(E)}$ we have $f\left(\frac{25}{11}\right)=10+f\left(\frac{1}{11}\right)=10-11=-1$ .

Thus, our answer is $\boxed{\textbf{(E)} \frac{25}{11}}$ ~JHawk0224

Solution 3 (Deeper)

Consider the rational $\frac{a}{b}$ , for $a,b$ integers. We have $f(a)=f\left(\frac{a}{b}\cdot b\right)=f\left(\frac{a}{b}\right)+f(b)$ . So $f\left(\frac{a}{b}\right)=f(a)-f(b)$ . Let $p$ be a prime. Notice that $f(p^k)=kf(p)$ . And $f(p)=[$ . So if $a=p_1^{a_1}p_2^{a_2}\cdots p_k^{a_k}$ , $f(a)=a_1p_1+a_2p_2+....+a_kp_k$ . We simply need this to be greater than what we have for $f(b)$ . Notice that for answer choices $A,B,C,$ and $D$ , the numerator $(a)$ has less prime factors than the denominator, and so none of these work (upon checking). Therefore, the answer is $\boxed{\textbf{(E)}}$ .

~yofro

Video Solution by Hawk Math

https://www.youtube.com/watch?v=dvlTA8Ncp58

Video Solution by Punxsutawney Phil

https://youtu.be/8gGcj95rlWY

Video Solution by OmegaLearn (Using Functions and manipulations)

https://youtu.be/aGv99CLzguE

~ pi_is_3.14

2021 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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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.

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