6. A busy bee buzzes between the cells of a large honeycomb made up of a plane of tessellated hexagons. A flight of length $n$ consists of picking any of the six neighbouring cells and flying to the $n^{\text {th }}$ cell in that direction. After consecutive flights of lengths $n=N, N-1, \ldots, 2,1$, the bee finds that it has returned to its starting location. For which values of $N$ is this possible?
SolUtion
If $N=1$ or $N=2$, the bee cannot return to its starting position.
If $N=3$ or $N=4$, the bee can return by going up and down only. In the first case, it goes 3 up, then 2 and 1 down. In the second case it goes 4 up, then 3 and 2 down, then 1 up.
If $N=5$ or $N=6$, the bee can return by forming an equilateral triangle. In the first case, it goes 5 in one direction, 4 and later 1 in another direction and 3 and 2 in the third direction. In the second case, it goes 6 and later 1 in one direction, 5 and later 2 in another direction and 4 and 3 in the third direction.

All further cases for $N$ can be reduced into a case $N-4$ by going up $N$, down $N-1$ and $N-2$, then up $N-3$, which returns it to its starting position. This process repeated will bring it down to one of the cases for $N$ being $3,4,5$ or 6 , which can be achieved.

Therefore, for all $N \geq 3$ the bee can return to its starting position.
AlternatiVe
As in the first solution, we check small cases and include that for $N=7$ or $N=8$, the bee can return by going up, down, down, up to start and then finishing using the results for $N=3$ and $N=4$ respectively.
All further cases for $N$ can be reduced into a case $N-6$ by going in the three directions $N$ and $N-5, N-1$ and $N-4, N-2$ and $N-3$, which returns it to its starting position by forming an equilateral triangle. This process repeated will bring it down to one of the cases for $N$ being 3 , $4,5,6,7$ or 8 which can be achieved.

Therefore, for all $N \geq 3$ the bee can return to its starting position.
AlternatiVe
Define the six hexagons adjacent to the central hexagon as the Ring.
If $N=1$ or $N=2$, the bee cannot return to its starting position, but the bee can return from the Ring to the starting position if there are 1 or 2 moves remaining.
Any pair of consecutive moves can result in a move by a single hexagon by reversing the direction for the second move.

Therefore, for any $N \geq 3$, the bee can return to its starting position by the following process. Move to the Ring after two moves. If there are still more than two moves remaining, move to an adjacent cell in the Ring in the next two moves, repeating this until there are one or two moves remaining. Return from the Ring to the starting position in the last one or two moves depending on the parity of $N$.