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

Revision as of 16:33, 23 August 2009

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 $f(600)$ ?

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

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

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

@@ Line 5: / Line 5: @@
 == Solution ==
-<math>f(500\cdot\frac65) = \frac3{\frac65} = \frac52</math>, so the answer is <math>\mathrm{C}</math>.
+<math>f(500\cdot\frac65) = \frac3{\frac65} = \frac52</math>, so the answer is <math>\boxed{\mathrm{C}}</math>.
 == See Also ==
 {{AMC12 box|year=2001|num-b=8|num-a=10}}