Difference between revisions of "2001 AIME II Problems/Problem 8"
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== Problem == | == Problem == | ||
| − | A certain function <math>f</math> has the properties that <math>f(3x) = 3f(x)</math> for all positive real values of <math>x</math>, and that <math>f(x) = 1 - \mid x - 2 \mid</math> for <math>1\leq x \leq 3</math>. Find the smallest <math>x</math> for which <math>f(x) = f(2001)</math>. | + | A certain [[function]] <math>f</math> has the properties that <math>f(3x) = 3f(x)</math> for all positive real values of <math>x</math>, and that <math>f(x) = 1 - \mid x - 2 \mid</math> for <math>1\leq x \leq 3</math>. Find the smallest <math>x</math> for which <math>f(x) = f(2001)</math>. |
== Solution == | == Solution == | ||
| − | {{ | + | Iterating the condition <math>f(3x) = 3f(x)</math>, we find that <math>f(x) = 3^kf\left(\frac{x}{3^k}\right)</math> for positive integers <math>k</math>. We know the definition of <math>f(x)</math> from <math>1 \le x \le 3</math>, so we would like to express <math>f(2001) = 3^kf\left(\frac{2001}{3^k}\right),\ 1 \le \frac{2001}{3^k} \le 3 \Longrightarrow k = 6</math>. Indeed, |
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| + | <cmath>f(2001) = 729\left[1 - \left| \frac{2001}{729} - 2\right|\right] = 186.</cmath> | ||
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| + | We now need the smallest <math>x</math> such that <math>f(x) = 3^kf\left(\frac{x}{3^k}\right) = 186</math>. The [[range]] of <math>f(x),\ 1 \le x \le 3</math>, is <math>0 \le f(x) \le 1</math>. Then <math>0 \le 186 = 3^kf\left(\frac{x}{3^k}\right) \le 3^k</math>, and the smallest value of <math>k</math> is <math>k = 5</math>. Then, | ||
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| + | <cmath>186 = 243\left[1 - \left| \frac{x}{243} - 2\right|\right] \Longrightarrow x = \pm 57 + 2 | ||
| + | \cdot 243</cmath> | ||
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| + | We want the smaller value of <math>x = \boxed{429}</math>. | ||
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| + | An alternative approach is to consider the graph of <math>f(x)</math>, which repeats every power of <math>3</math>, and resembles the section from <math>1 \le x \le 3</math> expanded by a factor of <math>3</math>. | ||
== See also == | == See also == | ||
{{AIME box|year=2001|n=II|num-b=7|num-a=9}} | {{AIME box|year=2001|n=II|num-b=7|num-a=9}} | ||
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| + | [[Category:Intermediate Algebra Problems]] | ||
Revision as of 15:00, 26 July 2008
Problem
A certain function 
has the properties that 
for all positive real values of 
, and that 
for 
. Find the smallest for which 
.
Solution
Iterating the condition , we find that 
for positive integers 
. We know the definition of 
from 
, so we would like to express 
. Indeed,
![\[f(2001) = 729\left[1 - \left| \frac{2001}{729} - 2\right|\right] = 186.\]](http://latex.artofproblemsolving.com/e/1/2/e129ad258113f974ab9e842efa3433c64a365217.png)
We now need the smallest such that 
. The range of 
, is 
. Then 
, and the smallest value of is 
. Then,
![\[186 = 243\left[1 - \left| \frac{x}{243} - 2\right|\right] \Longrightarrow x = \pm 57 + 2 \cdot 243\]](http://latex.artofproblemsolving.com/d/a/6/da615f6aa053bc14dd7296ad251d21ce42a5e06d.png)
We want the smaller value of 
.
An alternative approach is to consider the graph of , which repeats every power of 
, and resembles the section from expanded by a factor of .
See also
| 2001 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 7 |
Followed by Problem 9 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
