2012 AMC 12A Problems/Problem 25

Revision as of 08:13, 19 May 2012 by Yvain (talk | contribs) (Solution)

Problem

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.

$\textbf{(A)}\ 30\qquad\textbf{(B)}\ 31\qquad\textbf{(C)}\ 32\qquad\textbf{(D)}\ 62\qquad\textbf{(E)}\64$ (Error compiling LaTeX. Unknown error_msg)

Solution

C: n = 32

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