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 $\frac{500 \cdot 3}{600} = \boxed{\textbf{(C) } \frac52}$ .

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 $f(y)=\boxed{\textbf{(C) } \frac52}$ .

2001 AMC 12 (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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All AMC 12 Problems and Solutions

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2001 AMC 12 Problems/Problem 9

Contents

Problem

Solution 1

Solution 2

Solution 3

See Also