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 | ||
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