Difference between revisions of "2001 AIME II Problems/Problem 8"
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<cmath>f(2001) = 729\left[1 - \left| \frac{2001}{729} - 2\right|\right] = 186.</cmath> | <cmath>f(2001) = 729\left[1 - \left| \frac{2001}{729} - 2\right|\right] = 186.</cmath> | ||
| − | 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>. | + | 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>. So when <math>1 \le \frac{x}{3^k} \le 3</math>, we have <math>0 \le f\left(\frac{x}{3^k}\right) = \frac{186}{3^k} \le 1</math>. Multiplying by <math>3^k</math>: <math>0 \le 186 \le 3^k</math>, so the smallest value of <math>k</math> is <math>k = 5</math>. Then, |
| − | <cmath>186 = 243\left[1 - \left| \frac{x}{243} - 2\right|\right] \Longrightarrow x = \pm 57 + 2 | + | <cmath>186 = {3^5}f\left(\frac{x}{3^5}\right).</cmath> |
| − | \cdot 243</cmath> | + | |
| + | Because we forced <math>1 \le \frac{x}{3^5} \le 3</math>, so | ||
| + | |||
| + | <cmath>186 = {3^5}f\left(\frac{x}{3^5}\right) = 243\left[1 - \left| \frac{x}{243} - 2\right|\right] \Longrightarrow x = \pm 57 + 2 | ||
| + | \cdot 243.</cmath> | ||
We want the smaller value of <math>x = \boxed{429}</math>. | We want the smaller value of <math>x = \boxed{429}</math>. | ||
| − | An alternative approach is to consider the graph of <math>f(x)</math>, which | + | An alternative approach is to consider the graph of <math>f(x)</math>, which iterates every power of <math>3</math>, and resembles the section from <math>1 \le x \le 3</math> dilated by a factor of <math>3</math> each iteration. |
== See also == | == See also == | ||
Revision as of 11:32, 30 October 2016
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 
. So when 
, we have 
. Multiplying by 
: 
, so the smallest value of is 
. Then,
![\[186 = {3^5}f\left(\frac{x}{3^5}\right).\]](http://latex.artofproblemsolving.com/0/6/f/06f2b37921ee653205a539b6c9e2ab0be7c9a924.png)
Because we forced 
, so
![\[186 = {3^5}f\left(\frac{x}{3^5}\right) = 243\left[1 - \left| \frac{x}{243} - 2\right|\right] \Longrightarrow x = \pm 57 + 2 \cdot 243.\]](http://latex.artofproblemsolving.com/9/1/c/91c437fd96b60bcc9002ced02a93d102b03ce54a.png)
We want the smaller value of 
.
An alternative approach is to consider the graph of , which iterates every power of 
, and resembles the section from dilated by a factor of each iteration.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
