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

Revision as of 05:54, 6 June 2025

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 $y = \dfrac{6}{5}$ in the given equation, we get $f(600) = f(500\cdot\frac{6}{5}) = \frac{3}{\left(\frac{6}{5}\right)} = \boxed{\textbf{(C) } \frac{5}{2}}$ .

Solution 2a

Using the given equation, we obtain $f(xy) = f(yx) \iff \frac{f(x)}{y} = \frac{f(y)}{x} \iff xf(x) = yf(y),$ for all positive $x$ and $y$ . This means $xf(x)$ is constant, i.e. $xf(x) = k$ for some constant $k$ , and so the function $f$ must be of the form $f(x) = \frac{k}{x}$ . We can now compute $k$ using the given value of $f$ : $f(500) = 3 \iff \frac{k}{500} = 3 \iff k = 1500,$ which yields $f(600) = \frac{1500}{600} = \boxed{\textbf{(C) } \frac{5}{2}}.$

Solution 2b

Having determined that $f(x) = \frac{k}{x}$ for some constant $k$ , as in Solution 2a, an alternative way to finish the problem is to directly calculate: $f(600) = \frac{k}{600} = \frac{5}{6}\cdot\frac{k}{500} = \frac{5}{6}f(500) = \frac{5}{6} \cdot 3 = \boxed{\textbf{(C) } \frac{5}{2}}.$

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.

@@ Line 5: / Line 5: @@
 == 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>.
+Letting <math>x = 500</math> and <math>y = \dfrac{6}{5}</math> in the given equation, we get <math>f(600) = f(500\cdot\frac{6}{5}) = \frac{3}{\left(\frac{6}{5}\right)} = \boxed{\textbf{(C) } \frac{5}{2}}</math>.
-== Solution 2 ==
+== Solution 2a ==
-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>.
+Using the given equation, we obtain <cmath>f(xy) = f(yx) \iff \frac{f(x)}{y} = \frac{f(y)}{x} \iff xf(x) = yf(y),</cmath> for all positive <math>x</math> and <math>y</math>. This means <math>xf(x)</math> is constant, i.e. <math>xf(x) = k</math> for some constant <math>k</math>, and so the function <math>f</math> must be of the form <math>f(x) = \frac{k}{x}</math>. We can now compute <math>k</math> using the given value of <math>f</math>: <cmath>f(500) = 3 \iff \frac{k}{500} = 3 \iff k = 1500,</cmath> which yields <cmath>f(600) = \frac{1500}{600} = \boxed{\textbf{(C) } \frac{5}{2}}.</cmath>
-==Solution 3==
+== Solution 2b ==
-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>.
+Having determined that <math>f(x) = \frac{k}{x}</math> for some constant <math>k</math>, as in Solution 2a, an alternative way to finish the problem is to directly calculate: <cmath>f(600) = \frac{k}{600} = \frac{5}{6}\cdot\frac{k}{500} = \frac{5}{6}f(500) = \frac{5}{6} \cdot 3 = \boxed{\textbf{(C) } \frac{5}{2}}.</cmath>
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