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

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== Problem ==
 
== 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>. If <math>f(500) =3</math>, what is the value of <math>f(600)</math>?
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<!-- 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>
 
<math>(\mathrm{A})\ 1 \qquad (\mathrm{B})\ 2 \qquad (\mathrm{C})\ \frac52 \qquad (\mathrm{D})\ 3 \qquad (\mathrm{E})\ \frac{18}5</math>
  
== Solution ==
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== Solution 1 ==
<math>f(500\cdot\frac65) = \frac3{\frac65} = \frac52</math>, so the answer is <math>\boxed{\mathrm{C}}</math>.
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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>.
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== Solution 2 ==
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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} = \frac52</math>.
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==Solution 3==
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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)=\frac{5}{2}</math>.
  
 
== See Also ==
 
== See Also ==
  
 
{{AMC12 box|year=2001|num-b=8|num-a=10}}
 
{{AMC12 box|year=2001|num-b=8|num-a=10}}
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{{MAA Notice}}

Latest revision as of 21:18, 10 October 2020

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 (ProblemsAnswer KeyResources)
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

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