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Difference between revisions of "2001 AMC 12 Problems/Problem 9"
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== Solution 1 ==
 
== Solution 1 ==
−
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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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 ==
 
== 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} = \frac52</math>.
+
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==
 
==Solution 3==
−
Note that the equation given above is symmetric, so we have xf(x)=yf(y). Plugging in x=500 and y=600 gives f(y)=5/2
+
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 ==
 
== See Also ==
  
 
{{AMC12 box|year=2001|num-b=8|num-a=10}}
 
{{AMC12 box|year=2001|num-b=8|num-a=10}}
 
{{MAA Notice}}
 
{{MAA Notice}}

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

See Also

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

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