We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
73
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

12 Online Users

avatar