5. Silvia has 2006 identical triangular tiles and wants to arrange them all on the table without overlapping and in such a way that each tile shares exactly two sides with two other tiles. Can she succeed in her intention? Could she have succeeded last year, when she had 2005 tiles?

(A) It is impossible in both cases.

(B) It is possible with 2005 tiles, but not with 2006.

(C) It is possible with 2006 tiles, but not with 2005.

(D) In these two cases it is possible, but among the numbers greater than 12 there is at least one for which it is not possible.

(E) It is possible for all numbers of tiles greater than 12.
5. The answer is $\mathbf{( C )}$. The text asks for which $n$ it is possible to arrange $n$ tiles in a closed sequence where each tile has two adjacent tiles. Simple trials show that the first cases where this is possible are $n=6$ (an hexagon is formed where all tiles share a vertex), $n=12$ (a chain of tiles is formed with a triangular hole in the center, which is the difference between figures $F_{1}$ and $F_{0}$ of the next problem), $n=14$ (a chain of tiles is formed with a hole that is the union of two triangles). Focusing on the "hole," the region left free by the chain, it is immediately clear that if it is chosen to be a triangle of side $k$, exactly $6(k+1)$ tiles are needed to surround it; moreover, it is not difficult to convince oneself that if it is enlarged by a single triangular space, exactly two more tiles are needed to surround it, provided it is not enlarged in a non-convex part of its boundary. Therefore, by enlarging by zero, one, or two triangles a triangular hole of side $k-1$, regions are obtained that are surrounded by $6 k, 6 k+2$, and $6 k+4$ tiles, respectively. By choosing $k$ among the integers greater than 1, it follows that all even $n \geq 12$ admit a solution.

To see that solutions with odd $n$ are not possible, suppose one exists and imagine constructing it on an infinite triangular chessboard that has the triangles alternately colored white and black (so that adjacent triangles always have different colors). Two consecutive tiles will necessarily be placed on triangles of different colors, so the sequence of colors associated with the different tiles must be an alternating sequence of colors, and it must be closed, thus showing the same number of whites and blacks. Obviously, if $n$ is odd, this is not possible.