Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2001 AMC 12 Problems/Problem 9"

m (→‎Solution 3)
(→‎Solution 3)
 
(2 intermediate revisions by the same user not shown)
Line 5: Line 5:
  
 
== Solution 1 ==
 
== Solution 1 ==
−
Letting <math>x = 500</math> and <math>y = \dfrac65</math> in the given equation, we get <math>f(500\cdot\frac65) = \frac3{\frac65} = \frac52</math>, or <math>f(600) = \boxed{\textbf{C } \frac52}</math>.
+
Letting <math>x = 500</math> and <math>y = \dfrac65</math> in the given equation, we get <math>f(500\cdot\frac65) = \frac3{\frac65} = \frac52</math>, or <math>f(600) = \boxed{\textbf{(C) } \frac52}</math>.
  
 
== Solution 2 ==
 
== Solution 2 ==
−
The only function that satisfies the given condition is <math>y = \frac{k}{x}</math>, for some constant <math>k</math>. Thus, the answer is <math>\frac{500 \cdot 3}{600} = \frac52</math>.
+
The only function that satisfies the given condition is <math>y = \frac{k}{x}</math>, for some constant <math>k</math>. Thus, the answer is <math>\frac{500 \cdot 3}{600} = \boxed{\textbf{(C) } \frac52}</math>.
 +
 
 
==Solution 3==
 
==Solution 3==
−
Note that the equation given above is symmetric, so we have <math>x \cdot f(x)=y \cdot f(y)</math>. Plugging in <math>x=500</math> and <math>y=600</math> gives <math>f(y)=\frac{5}{2}</math>.
+
Note that the equation given above is symmetric, so we have <math>x \cdot f(x)=y \cdot f(y)</math>. Plugging in <math>x=500</math> and <math>y=600</math> gives <math>f(y)=\boxed{\textbf{(C) } \frac52}</math>.
  
 
== See Also ==
 
== See Also ==

Latest revision as of 23:33, 29 December 2023

Problem

Let $f$ be a function satisfying $f(xy) = \frac{f(x)}y$ for all positive real numbers $x$ and $y$. If $f(500) =3$, what is the value of $f(600)$?

$(\mathrm{A})\ 1 \qquad (\mathrm{B})\ 2 \qquad (\mathrm{C})\ \frac52 \qquad (\mathrm{D})\ 3 \qquad (\mathrm{E})\ \frac{18}5$

Solution 1

Letting $x = 500$ and $y = \dfrac65$ in the given equation, we get $f(500\cdot\frac65) = \frac3{\frac65} = \frac52$, or $f(600) = \boxed{\textbf{(C) } \frac52}$.

Solution 2

The only function that satisfies the given condition is $y = \frac{k}{x}$, for some constant $k$. Thus, the answer is $\frac{500 \cdot 3}{600} = \boxed{\textbf{(C) } \frac52}$.

Solution 3

Note that the equation given above is symmetric, so we have $x \cdot f(x)=y \cdot f(y)$. Plugging in $x=500$ and $y=600$ gives $f(y)=\boxed{\textbf{(C) } \frac52}$.

See Also

2001 AMC 12 (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2001_AMC_12_Problems/Problem_9&oldid=208843"