2001 AIME II Problems/Problem 8
Contents
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 at each iteration.
Solution 2 (Graphing)
First, we start by graphing the function when , which consists of the lines and that intersect at . Similarly, using , we get a dilation of our initial figure by a factor of 3 for the next interval and so on. Observe that the intersection of two lines always has coordinates where for some . First, we compute . The nearest intersection point is when . Therefore, we can safely assume that is somewhere on the line with a slope of that intersects at that nearest point. Using the fact that the slope of the line is , we compute . However, we want the minimum value such that and we see that there is another intersection point on the left which has a , namely . Therefore, we want the point that lies on the line with slope that intersects this point. Once again, since the slope of the line is , we get .
Solution 3 (Complete Bash but FAST)
We evaluate the first few terms of f(x) to try to find a pattern.
F(1)=0 F(2)=1 F(3)=0 F(4)=1 F(5) = 3(F()) = 2
That doesn‘t seem to be getting us anywhere. We notice what we did with f(5) will probably work with f(2001).
From here, we can evaluate f(2001) = when we plug in into . So all we need to find is the least number, let‘s call it, say y such that f(y)=186.
Repeating the same process we did before with f(2001),
Notice that we stopped at because is inside the range of , which is [0,1]. Now, f(y/243) = 186/243. Setting , we get 2 solutions for x: and .
Now, the problem asks for the smallest solution, so we obviously choose 429/243 as the solution for 243f(y/243) because it is smaller.
We found that , and solving this equation gives our answer
~MathCosine
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.