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

Revision as of 15:29, 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 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

Video Solution by Punxsutawney Phil

https://youtu.be/8gGcj95rlWY

@@ Line 20: / Line 20: @@
 <cmath>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</cmath>
 <cmath>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</cmath>
-In <math>A)</math> we have <math>f\left(\frac{17}{32}\right)=17+5f\left(\frac{1}{2}\right)=17-5(2)=7</math>.\\
+In <math>\textbf{(A)}</math> we have <math>f\left(\frac{17}{32}\right)=17+5f\left(\frac{1}{2}\right)=17-5(2)=7</math>.
-In <math>B)</math> we have <math>f\left(\frac{11}{16}\right)=11+4f\left(\frac{1}{2}\right)=11-4(2)=3</math>.\\
+In <math>\textbf{(B)}</math> we have <math>f\left(\frac{11}{16}\right)=11+4f\left(\frac{1}{2}\right)=11-4(2)=3</math>.
-In <math>C)</math> we have <math>f\left(\frac{7}{9}\right)=7+2f\left(\frac{1}{3}\right)=7-2(3)=1</math>.\\
+In <math>\textbf{(C)}</math> we have <math>f\left(\frac{7}{9}\right)=7+2f\left(\frac{1}{3}\right)=7-2(3)=1</math>.
-In <math>D)</math> we have <math>f\left(\frac{7}{6}\right)=7+f\left(\frac{1}{2}\right)+f\left(\frac{1}{3}\right)=7-2-3=2</math>.\\
+In <math>\textbf{(D)}</math> we have <math>f\left(\frac{7}{6}\right)=7+f\left(\frac{1}{2}\right)+f\left(\frac{1}{3}\right)=7-2-3=2</math>.
-In <math>E)</math> we have <math>f\left(\frac{25}{11}\right)=10+f\left(\frac{1}{11}\right)=10-11=-1</math>.\\
+In <math>\textbf{(E)}</math> we have <math>f\left(\frac{25}{11}\right)=10+f\left(\frac{1}{11}\right)=10-11=-1</math>.
 Thus, our answer is <math>\boxed{\textbf{(E)} \frac{25}{11}}</math>
 ~JHawk0224

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

Page

Toolbox

Search