Task $1 / 83$ On a corner of a cube sits a mathematically trained caterpillar. She wants to try all the paths along the edges that lead to the diametrically opposite corner, without visiting any corner twice.

To crawl along an edge, she needs one day; at night she rests. Upon reaching the destination, she slides back to the starting point along the cube's diagonal during the following night, to begin a new path the next morning.

She starts on the morning of January 1, 1983. On the morning of which day will she be back at the starting point, after having tried all the paths?
The cube has vertices $A(0 ; 0,0), B(1 ; 0,0)$, $C(1 ; 1 ; 0), D(0 ; 1 ; 0), E(0 ; 0 ; 1), F(1 ; 0,1), G(1 ; 1 ; 1)$, and $H(0 ; 1 ; 1)$ in a right-angled Cartesian coordinate system. Due to symmetry, it suffices to first determine the number of paths that start with the edge $A B$, as the same number applies to paths starting with $A C$ or $A E$.

From $B$, the continuation is possible on two "symmetrical" paths: $B C$ or $B F$. We consider $B C$. The following paths exist:

$$
\begin{array}{rl}
A B-B C-C G & 3 \text { edges } \\
A B-B C-C D-D H-H G & 5 \text { edges } \\
A B-B C-C D-D H-H E-E F-F G & 7 \text { edges }
\end{array}
$$

Thus, there are 6 paths with 3 edges, 6 paths with 5 edges, and 6 paths with 7 edges, making a total of $6(3+5+7)=90$ edges that the caterpillar must crawl through. This takes 90 days; on the morning of the 91st day, which is April 1, 1983, the caterpillar, having completed its work, is back at the starting point.