+0  
 
0
127
1
avatar

For each \(x\) in \([0,1]\), define \(\begin{cases} f(x) = 2x, \qquad\qquad \mathrm{if} \quad 0 \leq x \leq \frac{1}{2};\\ f(x) = 2-2x, \qquad \mathrm{if} \quad \frac{1}{2} < x \leq 1. \end{cases}\) Let \(f^{[2]}(x) = f(f(x))\), and \(f^{[n + 1]}(x) = f^{[n]}(f(x))\) for each integer \(n \geq 2\). Then the number of values of \(x\) in \([0,1]\) for which \(f^{[2005]}(x) = \frac {1}{2}\) can be expressed in the form \(p^a,\) where \(p\) is a prime and \(a\) is a positive integer. Find \(p + a.\)

 Apr 10, 2019
 #1
avatar
0

The answer is 4010 + 2 = 4012.

 Nov 29, 2019

6 Online Users

avatar