2001 AMC 12 Problems/Problem 9: Difference between revisions

Revision as of 19:54, 24 September 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

no

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: $f(600) = \frac{k}{600} = \frac{5}{6}\cdot\frac{k}{500} = \frac{5}{6}f(500) = \frac{5}{6} \cdot 3 = \boxed{\textbf{(C) } \frac{5}{2}}.$

Solution 3 (educated guessing)

Notice that $500$ and $600$ both have $100$ in common, so set $x = 100$ and $y = 5$ . Thus, we get $f(500) = \frac{f(100)}{5} = 3 \iff f(100) = 15.$ Then, since we know $f(600) = f(100\cdot6),$ we get $f(600) = \frac{f(100)}{6} = \frac{15}{6} = \boxed{\textbf{(C) } \frac{5}{2}}.$

2001 AMC 12 (Problems • Answer Key • Resources)
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 == Solution 2a ==
-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>
+no
 == Solution 2b ==

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