AMC 12 2012a Problem 25

Let $f(x)=|2\{x\}-1|$ where $\{x\}$ denotes the fractional part of $x$. The number $n$ is the smallest positive integer such that the equation $$nf(xf(x))=x$$ has at least $2012$ real solutions. What is $n$? '''Note:''' the fractional part of $x$ is a real number $y=\{x\}$ such that $0\le y<1$ and $x-y$ is an integer.

\textbf{(A)}\ 30 \textbf{(B)}\ 31 \textbf{(C)}\ 32 \textbf{(D)}\ 62 \textbf{(E)}\ 64