13. (LUX 2) Let $E$ be the set of $1983^{3}$ points of the space $\mathbb{R}^{3}$ all three of whose coordinates are integers between 0 and 1982 (including 0 and 1982). A coloring of $E$ is a map from $E$ to the set \{red, blue \}. How many colorings of $E$ are there satisfying the following property: The number of red vertices among the 8 vertices of any right-angled parallelepiped is a multiple of 4 ?
13. Given any coloring of the $3 \times 1983-2$ points of the axes, we prove that there is a unique coloring of $E$ having the given property and extending this coloring. The first thing to notice is that given any rectangle $R_{1}$ parallel to a coordinate plane and whose edges are parallel to the axes, there is an even number $r_{1}$ of red vertices on $R_{1}$. Indeed, let $R_{2}$ and $R_{3}$ be two other rectangles that are translated from $R_{1}$ orthogonally to $R_{1}$ and let $r_{2}, r_{3}$ be the numbers of red vertices on $R_{2}$ and $R_{3}$ respectively. Then $r_{1}+r_{2}$, $r_{1}+r_{3}$, and $r_{2}+r_{3}$ are multiples of 4 , so $r_{1}=\left(r_{1}+r_{2}+r_{1}+r_{3}-r_{2}-r_{3}\right) / 2$ is even.
Since any point of a coordinate plane is a vertex of a rectangle whose remaining three vertices lie on the corresponding axes, this determines uniquely the coloring of the coordinate planes. Similarly, the coloring of the inner points of the parallelepiped is completely determined. The solution is hence $2^{3 \times 1983-2}=2^{5947}$.