Looks easy but it Isn't

Let $f(x) = | \lfloor x \rfloor - 2x|$ . It is given that $| \lfloor x \rfloor - 2x| = 4$ . Since the RHS is an integer and $\lfloor x \rfloor$ is also an integer, $2x$ must be an integer. Therefore, $x = n$ or $x = n +0.5$ , where $n$ is an integer.

For $x = n$ ,

$f(x) = | n - 2n | = | - n | =4 \quad \Rightarrow n = \pm 4 \quad x = n = \pm 4$

For $x = n+0.5$ ,

$f(x) = | n - 2(n+0.5) | = | - n-1 | =4 \quad \Rightarrow n = 3$ or $-5 \quad x = n+0.5 = 3.5$ or $-4.5$

There are $a = 4$ solutions. And the answer is $-4.5 \times -4 \times 4 \times 3.5 \times 4 = \boxed{1008}$ .

Simple algebra that $\lfloor x \rfloor = 2x+4 \text{ or } 2x-4$

We see that $2x+4, 2x-4 \in \mathbb{Z}$ , $x$ must be in the form of $\displaystyle \frac{k}{2}$ for any integer $k$ . (i.e. integer or integer + 0.5)

Case 1: $\lfloor x \rfloor = 2x+4$

$2x+4 \leq x < 2x+5$

Which gives $x = -4, -4.5$ .

Case 2: $\lfloor x \rfloor = 2x-4$

$2x-4 \leq x < 2x-3$

Which gives $x = 4, 3.5$

Therefore, the solutions are $\boxed{x = -4.5, -4, 3.5, 4}$ ~~~