The rook, standing on the surface of the checkered cube, beats the cells, located in the same row as well as on the
continuations of this series through one or even several edges. (The picture shows an example for a $4 \times 4 \times 4$ cube,visible cells that some beat the rook, shaded gray.) What is the largest number do not beat each other rooks can be placed on the surface of the cube $50 \times 50 \times 50$?
$75$ rooks.
Suppose that left and right face $=x$, back and forth face $=y$, up and down face $=z$.
We know that that number of rooks on every two of $x,y,z$ are less than or equal to $50$ so we have$:$
$x+y+z=75 \begin{cases}
x+y \le 50\\
x+z \le 50\\
z+y \le 50\\
\end{cases}$
