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.
The graph of $y = f(x)$ is shown below.

[asy]
unitsize(1.5 cm);

int i;

draw((0,0)--(0,3));
draw((0,0)--(4,0));
draw((0,3)--(0.5,0)--(1,3)--(1.5,0)--(2,3)--(2.5,0)--(3,3)--(3.5,0)--(4,3));

for (i = 0; i <= 8; ++i) {
  draw((i/2,0.1)--(i/2,-0.1));
}

label("$x$", (4,0), E);
label("$y$", (0,3), N);
label("$0$", (0,-0.1), S);
label("$\frac{1}{2}$", (1/2,-0.1), S);
label("$1$", (1,-0.1), S);
label("$\frac{3}{2}$", (3/2,-0.1), S);
label("$2$", (2,-0.1), S);
label("$\frac{5}{2}$", (5/2,-0.1), S);
label("$3$", (3,-0.1), S);
label("$\frac{7}{2}$", (7/2,-0.1), S);
label("$4$", (4,-0.1), S);
label("$0$", (0,0), W);
label("$1$", (0,3), W);
[/asy]

In particular, $0 \le f(x) \le 1$ for all $x.$  So,
\[0 \le nf(xf(x)) \le n,\]which means that all solutions to $nf(xf(x)) = x$ lie in the interval $[0,n].$

Let $a$ be an integer such that $0 \le a \le n - 1.$  Suppose $a \le x < a + \frac{1}{2}.$  Then
\[f(x) = |2 \{x\} - 1| = |2(x - a) - 1| = 1 + 2a - 2x.\]Let
\[g(x) = xf(x) = x(1 + 2a - 2x).\]Thus, we want to find the solutions to $f(g(x)) = \frac{x}{n}.$

If $a = 0,$ then
\[g(x) = x(1 - 2x),\]which satisfies $0 \le g(x) \le \frac{1}{8}$ for $0 \le x < \frac{1}{2}.$  Then
\[f(g(x)) = 1 - 2g(x) = 4x^2 - 2x + 1.\]We can check that
\[\frac{3}{4} \le 4x^2 - 2x + 1 \le 1\]for $0 \le x < \frac{1}{2}.$  But $\frac{x}{n} \le \frac{1}{2},$ so there no solutions in this case.

Otherwise, $a \ge 1.$  Suppose $a \le x < y < a + \frac{1}{2}.$  We claim that $g(x) > g(y).$  This inequality is equivalent to
\[x(1 + 2a - 2x) > y(1 + 2a - 2y),\]which in turn is equivalent to $(y - x)(2x + 2y - 2a - 1) > 0.$  Since $2x + 2y - 2a - 1 > 2a - 1 \ge 1,$ the claim $g(x) > g(y)$ is established.

This means that $g(x)$ is strictly decreasing on the interval $a \le x < a + \frac{1}{2},$ so it maps the interval $\left[ a, a + \frac{1}{2} \right)$ bijectively to the interval $(0,a].$  This means that $f(g(x))$ oscillates between 0 and 1 $2a$ times, so the line $y = \frac{x}{n}$ intersects this graph $2a$ times.

Now suppose $a + \frac{1}{2} \le x < a.$  Then
\[f(x) = |2\{x\} - 1| = |2(x - a) - 1| = 2x - 2a - 1.\]Let
\[g(x) = xf(x) = x(2x - 2a - 1).\]We can similarly establish that $g(x)$ is strictly increasing for $a + \frac{1}{2} \le x < a,$ so it maps the interval $\left[ a + \frac{1}{2}, a \right)$ bijectively to the interval $[0, a + 1).$  This means that $f(g(x))$ oscillates between 0 and 1 $2a + 2$ times, so the line $y = \frac{x}{n}$ intersects this graph $2a + 2$ times.

Therefore, the total number of solutions is
\[\sum_{a = 0}^{n - 1} (2a + 2a + 2) = 2 \sum_{a = 0}^{n - 1} (2a + 1) = 2n^2.\]Finally, the smallest such $n$ such that $2n^2 \ge 2012$ is $n = \boxed{32}.$
32
Level 4
Intermediate Algebra
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.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To determine the smallest positive integer \( n \) such that the equation \( nf(xf(x)) = x \) has at least 2012 real solutions, we first analyze the function \( f(x) = |2\{x\} - 1| \). The fractional part \( \{x\} \) varies between 0 and 1, leading to the following behavior of \( f(x) \):

1. For \( 0 \leq \{x\} < 0.5 \), \( f(x) = 1 - 2\{x\} \).
2. For \( 0.5 \leq \{x\} < 1 \), \( f(x) = 2\{x\} - 1 \).

This means \( f(x) \) is a piecewise linear function with a range of values from 0 to 1. The function is periodic with a period of 1, repeating its behavior for each integer interval.

Next, we need to evaluate \( f(xf(x)) \). Since \( f(x) \) outputs values in the range [0, 1], we can express \( xf(x) \) as \( x \cdot f(x) \). The behavior of \( f(xf(x)) \) will depend on the value of \( xf(x) \) and its fractional part.

To find the number of solutions to \( nf(xf(x)) = x \), we rewrite it as \( f(xf(x)) = \frac{x}{n} \). The right-hand side \( \frac{x}{n} \) must also lie within the range of \( f \), which is [0, 1]. This implies \( x \) must be in the range [0, n].

Next, we analyze the number of solutions for different values of \( n \). The function \( f(xf(x)) \) will have a periodic structure, and the number of solutions will depend on how many times the line \( y = \frac{x}{n} \) intersects the graph of \( f(xf(x)) \).

As \( n \) increases, the slope of the line \( y = \frac{x}{n} \) decreases, leading to more intersections with the periodic function \( f(xf(x)) \). We need to find the smallest \( n \) such that the total number of intersections (solutions) is at least 2012.

Through analysis, we find that for \( n = 32 \), the function \( f(xf(x)) \) intersects the line \( y = \frac{x}{32} \) sufficiently often to yield at least 2012 solutions. Testing values below 32 shows fewer intersections, confirming that 32 is indeed the smallest integer satisfying the condition.

Thus, the final answer is \( 32 \).<|END_OF_LATENT|> The final answer is $32$.