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

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== Solution ==
 
== Solution ==
<math>f(500\cdot\frac65) = \frac3{\frac65} = \frac25</math>, so the answer is <math>\mathrm{C}</math>.
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<math>f(500\cdot\frac65) = \frac3{\frac65} = \frac52</math>, so the answer is <math>\mathrm{C}</math>.

Revision as of 00:51, 8 February 2009

Problem

Let $f$ be a function satisfying $f(xy) = \frac{f(x)}y$ for all postitive 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 $\mathrm{C}$.