Difference between revisions of "2001 AMC 12 Problems/Problem 9"
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(Another solution where basically you solve for f(100) then solve for f(600)) |
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== Solution 1 == | == Solution 1 == | ||
| − | Letting <math>x = 500</math> and <math>y = \ | + | 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>. |
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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> | ||
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| + | == Solution 2b == | ||
| + | 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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| + | == Solution 3 (educated guessing) == | ||
| + | Notice that <math>500</math> and <math>600</math> both have <math>100</math> in common, so set <math>x = 100</math> and <math>y = 5</math>. Thus, we get <cmath>f(500) = \frac{f(100)}{5} = 3 \iff f(100) = 15.</cmath> Then, since we know <cmath>f(600) = f(100\cdot6),</cmath> we get <cmath>f(600) = \frac{f(100)}{6} = \frac{15}{6} = \boxed{\textbf{(C) } \frac{5}{2}}.</cmath> | ||
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== See Also == | == See Also == | ||
Latest revision as of 17:27, 11 June 2025
Contents
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 
.
Solution 2a
Using the given equation, we obtain ![\[f(xy) = f(yx) \iff \frac{f(x)}{y} = \frac{f(y)}{x} \iff xf(x) = yf(y),\]](http://latex.artofproblemsolving.com/1/1/8/118997f0b9b964d35d762db6807ebc675b7fd04e.png)
for all positive and . This means 
is constant, i.e. 
for some constant 
, and so the function must be of the form 
. We can now compute using the given value of : ![\[f(500) = 3 \iff \frac{k}{500} = 3 \iff k = 1500,\]](http://latex.artofproblemsolving.com/6/f/a/6fa89b5c1e85f551917d4a3c23267c21a9e3a8a8.png)
which yields ![\[f(600) = \frac{1500}{600} = \boxed{\textbf{(C) } \frac{5}{2}}.\]](http://latex.artofproblemsolving.com/a/d/9/ad95b7fd2d99b640780cb8c1d2b56e8359e2ca15.png)
Solution 2b
Having determined that for some constant , as in Solution 2a, an alternative way to finish the problem is to directly calculate: ![\[f(600) = \frac{k}{600} = \frac{5}{6}\cdot\frac{k}{500} = \frac{5}{6}f(500) = \frac{5}{6} \cdot 3 = \boxed{\textbf{(C) } \frac{5}{2}}.\]](http://latex.artofproblemsolving.com/5/9/3/593dfc969b7ed3e4dba78178fc096befb77e6e42.png)
Solution 3 (educated guessing)
Notice that 
and 
both have 
in common, so set 
and 
. Thus, we get ![\[f(500) = \frac{f(100)}{5} = 3 \iff f(100) = 15.\]](http://latex.artofproblemsolving.com/1/e/d/1edbaff1591b5b25abd93a1adb3c73d528d81d24.png)
Then, since we know ![\[f(600) = f(100\cdot6),\]](http://latex.artofproblemsolving.com/9/b/b/9bb85348718d9c01e2a2904985c49a0fadf6d20d.png)
we get ![\[f(600) = \frac{f(100)}{6} = \frac{15}{6} = \boxed{\textbf{(C) } \frac{5}{2}}.\]](http://latex.artofproblemsolving.com/3/7/c/37cad113f7337d7285bb81df05aecc93265a9f0f.png)
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 | |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
