# 2001 AMC 12 Problems/Problem 9

## 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 $\frac{500 \cdot 3}{600} = \frac52$.

## 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 $f(y)=\frac{5}{2}$.

## See Also

 2001 AMC 12 (Problems • Answer Key • Resources) Preceded byProblem 8 Followed byProblem 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

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