Area of greatest integer function Question:

Find the area enclosed by the function:
  $$\left\lfloor\frac{\left|3x + 4y\right|}{5}\right\rfloor + \left\lfloor\frac{\left|4x - 3y\right|}{5}\right\rfloor = 3$$
  where $\lfloor\cdot\rfloor$ represents the greatest-integer function.

I believe the easiest way to solve this problem would be to graph the function. However, I was unable to do so. Please note that this was a question on an examination and there was no graphing equipment available.
 A: Maybe you remember that there is a right triangle with sides $3$, $4$, $5$. This then implies that
$$u:={4\over5}x-{3\over5}y,\qquad v:={3\over5}x+{4\over5}y$$
are orthonormal coordinates in the plane (see the figure). In terms of these coordinates the given equation reads
$$\bigl\lfloor |v|\bigr\rfloor+\bigl\lfloor |u|\bigr\rfloor=3\ .\tag{1}$$
Draw an $(u,v)$-plane with horizontal $u$-axis and vertical $v$-axis. Because of the absolute value signs the set $S$ of points $(u,v)$ satisfying $(1)$ is symmetric with respect to both axes. Therefore it is sufficient to determine the part of $S$ lying in the first quadrant. 
Now stare at the figure and find out which points in the first quadrant of the $(u,v)$-plane satisfy $\lfloor u\rfloor+\lfloor v\rfloor =3$.

A: We can separate it into cases :
$$\left\lfloor\frac{|3x+4y|}{5}\right\rfloor+\left\lfloor\frac{|4x-3y|}{5}\right\rfloor=3$$
$$\iff \left(\left\lfloor\frac{|3x+4y|}{5}\right\rfloor,\left\lfloor\frac{|4x-3y|}{5}\right\rfloor\right)=(0,3),(1,2),(2,1),(3,0)$$
Now note that 
$$\left\lfloor Z\right\rfloor=a\iff a\le Z\lt a+1$$
and that $\frac{|3X+4Y|}{5},\frac{|4X-3Y|}{5}$ represents the distance between a point $(X,Y)$ and the line $3x+4y=0, 4x-3y=0$ respectively. 
Since the two lines intersect perpendicularly, we have $16$ unit squares, and so the answer is $\color{red}{16}$.
