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>? | + | <!-- 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 == | + | == Solution 1 == |
− | <math>f(500\cdot\frac65) = \frac3{\frac65} = \frac52</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} = \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)=\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}} | ||
+ | {{MAA Notice}} |
Latest revision as of 21:18, 10 October 2020
Problem
Let be a function satisfying for all positive real numbers and . If , what is the value of ?
Solution 1
Letting and in the given equation, we get , or .
Solution 2
The only function that satisfies the given condition is , for some constant . Thus, the answer is .
Solution 3
Note that the equation given above is symmetric, so we have . Plugging in and gives .
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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