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

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== Problem ==
 
== Problem ==
Let <math>f</math> be a function satisfying <math>f(xy) = \frac{f(x)}y</math> for all postitive real numbers <math>x</math> and <math>y</math>, and <math>f(500) =3</math>. What is <math>f(600)</math>?
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Let <math>f</math> be a function satisfying <math>f(xy) = \frac{f(x)}y</math> for all positive real numbers <math>x</math> and <math>y</math>, and <math>f(500) =3</math>. What is <math>f(600)</math>?
  
 
<math>(\mathrm{A})\ 1 \qquad (\mathrm{B})\ 2 \qquad (\mathrm{C})\ \frac52 \qquad (\mathrm{D})\ 3 \qquad (\mathrm{E})\ \frac{18}5</math>
 
<math>(\mathrm{A})\ 1 \qquad (\mathrm{B})\ 2 \qquad (\mathrm{C})\ \frac52 \qquad (\mathrm{D})\ 3 \qquad (\mathrm{E})\ \frac{18}5</math>

Revision as of 01:52, 8 February 2009

Problem

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

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

Solution

$f(500\cdot\frac65) = \frac3{\frac65} = \frac52$, so the answer is $\mathrm{C}$.