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

Latest revision as of 23:33, 29 December 2023

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
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 1: / Line 1: @@
 == Problem ==
-Let <math>f</math> be a function satisfying <math>f(xy) = \frac{f(x)}y</math> for all positive real numbers <math>x</math> and <math>y</math>, and <math>f(500) =3</math>. What is <math>f(600)</math>?
+<!-- don't remove the following tag, for PoTW on the Wiki front page--><onlyinclude>Let <math>f</math> be a function satisfying <math>f(xy) = \frac{f(x)}y</math> for all positive real numbers <math>x</math> and <math>y</math>. If <math>f(500) =3</math>, what is the value of <math>f(600)</math>?<!-- don't remove the following tag, for PoTW on the Wiki front page--></onlyinclude>
 <math>(\mathrm{A})\ 1 \qquad (\mathrm{B})\ 2 \qquad (\mathrm{C})\ \frac52 \qquad (\mathrm{D})\ 3 \qquad (\mathrm{E})\ \frac{18}5</math>
-== Solution ==
+== Solution 1 ==
-<math>f(500\cdot\frac65) = \frac3{\frac65} = \frac52</math>, so the answer is <math>\mathrm{C}</math>.
+Letting <math>x = 500</math> and <math>y = \dfrac65</math> in the given equation, we get <math>f(500\cdot\frac65) = \frac3{\frac65} = \frac52</math>, or <math>f(600) = \boxed{\textbf{(C) } \frac52}</math>.
+== Solution 2 ==
+The only function that satisfies the given condition is <math>y = \frac{k}{x}</math>, for some constant <math>k</math>. Thus, the answer is <math>\frac{500 \cdot 3}{600} = \boxed{\textbf{(C) } \frac52}</math>.
+==Solution 3==
+Note that the equation given above is symmetric, so we have <math>x \cdot f(x)=y \cdot f(y)</math>. Plugging in <math>x=500</math> and <math>y=600</math> gives <math>f(y)=\boxed{\textbf{(C) } \frac52}</math>.
+== See Also ==
+{{AMC12 box|year=2001|num-b=8|num-a=10}}
+{{MAA Notice}}

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