, 2006

Solution:

The answer is $\mathbf{(C)}$. The text asks for which $n$ it is possible to join $n$ tiles in a closed sequence in which each tile has two other adjacent tiles. By making some simple trials one immediately sees that the first cases in which it is possible are $n=6$ (one obtains a hexagon in which the tiles all have a vertex in common), $n=12$ (one obtains a chain of tiles with a triangular hole in the center, which is then the difference between the figures $F_{1}$ and $F_{0}$ of the following problem), $n=14$ (one obtains a chain of tiles with a hole that is the union of two triangles).

Turning our attention to the "hole", the region without tiles left free by the chain, it is immediate that if it is chosen to be shaped like a triangle of side $k$, exactly $6(k+1)$ tiles are needed to surround it; moreover it is not hard to convince oneself that if it is enlarged by a single triangular space, exactly two more tiles are needed to surround it, provided one has not enlarged it in a non-convex zone of its boundary. So by enlarging a triangular hole of side $k-1$ by zero, one, or two triangles one obtains regions that are surrounded by $6k$, $6k+2$ and $6k+4$ tiles respectively. Since $k$ can be chosen among the integers greater than $1$, one obtains that all even $n \geq 12$ admit a solution.

To see that solutions with odd $n$ are not possible, it suffices to suppose that one exists and to imagine building it on an infinite triangular chessboard whose triangles are colored alternately white and black (so that two triangles adjacent along a side always have different color). Two consecutive tiles will necessarily lie on triangles of different color, so the sequence of colors associated with the various tiles must be an alternating sequence of colors, and it must be closed, hence it must show the same number of whites and of blacks. Obviously, if $n$ is odd this is not possible.