2014 AMC 12A Problems/Problem 24
Contents
[hide]Problem
Let , and for , let . For how many values of is ?
Solution 1
1. Draw the graph of by dividing the domain into three parts.
2. Apply the recursive rule a few times to find the pattern.
Note: is used to enlarge the difference, but the reasoning is the same.
3. Extrapolate to . Notice that the summits start away from and get closer each iteration, so they reach exactly at .
reaches at , then zigzags between and , hitting at every even , before leaving at .
This means that at all even where . This is a -integer odd-size range with even numbers at the endpoints, so just over half of the integers are even, or . (Revised by Flamedragon & Jason,C & emerald_block)
Solution 2
First, notice that the recursive rule moves the current value closer to . Upon reaching , it alternates between and . This means that exactly when (to reach in time) and is even (so ).
Casework each part of (where the expressions in the absolute values do not change sign): so even work. so even work. so even work.
Putting these together, all even where work. So, the answer is . ~revised by emerald_block
Solution 3
Note when = 0. This occurs when .
Then, repeating this process, we note , and hence .
Similarly, . Extrapolating this pattern, we must have , , . Then, drawing the graph of , we note for each of , , , there are three solutions. For , there is exactly solutions.
So, the total amount of solutions is
Video Solution by Richard Rusczyk
https://artofproblemsolving.com/videos/amc/2014amc12a/383
~ dolphin7
See Also
2014 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 23 |
Followed by Problem 25 |
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