How many solution with the equation $f_{2013}(x)=\frac{x}{2013}$ 
let $f(x)=f_{1}(x)=\mid \cos{(2\pi x)}\mid,f_{2}(x)=f(f_{1}(x))=\mid \cos{(2\pi (\mid\cos{(2\pi x)}\mid)}\mid$
  $f_{n}(x)=f(f_{n-1}(x))$,

Question:

How many solution with the equation
  $$f_{2013}(x)=\dfrac{x}{2013}$$

This problem is from china students ask me,But I consider sometimes,and I can't it,I hope someone can help.
My try: I find $$\cos{(2\pi x)}=\dfrac{x}{2013}$$

 A: The map 
$$
\varphi : [0,1] \to [0,1], \qquad \varphi(t) = |\cos(2\pi t)|
$$
is surjective and continuous. During the interval $[0,1/4]$, $\varphi$ goes from $1$ to $0$. Then during the interval $[1/4,1/2]$, $\varphi$ goes from $0$ to $1$. Again during $[1/2,3/4]$, $\varphi$ goes from $1$ to $0$, and finally during $[3/4,1]$, $\varphi$ goes back from $0$ to $1$. 
The line $y = x/2013$ must therefore cross the graph of $\varphi$ 4 times, namely the number of times the curve $(t,\varphi(t))$ goes from $\varphi(x_1) = 0$ to $\varphi(x_2) = 1$ or vice-versa.
If we iterate $\varphi$ to get $\varphi^2$, over each of those intervals of size $1/4$, $\varphi$ will go back and forth as explained above $4$ times, for a total of $16 = 4^2$ times. For $\varphi^3$, it will go back and forth $64 = 4^3$ times. 
The number of such $x$ is therefore $4^{2013}$. 
EDIT! Perhaps I should add : in the intervals where $\varphi$ goes from $0$ to $1$ (resp. $1$ to $0$), $\varphi$ is increasing (resp. decreasing), hence will intersect the line $y = x/2013$ precisely once. 
Hope that helps,
