2001 AMC 12 Problems/Problem 9: Difference between revisions

Revision as of 00:51, 8 February 2009

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 $f(600)$ ?

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

$f(500\cdot\frac65) = \frac3{\frac65} = \frac25$ , so the answer is $\mathrm{C}$ .

Revision as of 00:51, 8 February 2009 view source Duelist (talk \| contribs) editor 49 edits New page: == 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>. Wh...		Revision as of 00:51, 8 February 2009 view source Duelist (talk \| contribs) editor 49 edits →Solution Newer edit →
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	== Solution ==		== Solution ==
	<math>f(500\cdot\frac65) = \frac3{\~~frac{65}~~} = \frac25</math>, so the answer is <math>\mathrm{C}</math>.		<math>f(500\cdot\frac65) = \frac3{\frac65} = \frac25</math>, so the answer is <math>\mathrm{C}</math>.