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 - | + | 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-|x-2|</math> for <math>1\le x \le 3</math>. Find the smallest <math>x</math> for which <math>f(x) = f(2001)</math>. |
== Solution == | == Solution == |
Revision as of 11:34, 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,
We now need the smallest such that . The range of , is . So when , we have . Multiplying by : , so the smallest value of is . Then,
Because we forced , so
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.