Found problems: 80
You have a large number of congruent equilateral triangular tiles on a table and you want to fit $n$ of them together to make a convex equiangular hexagon (i.e. one whose interior angles are $120^o$) . Obviously, $n$ cannot be any positive integer. The first three feasible $n$ are $6, 10$ and $13$. Show that $12$ is not feasible but $14$ is.
In a checkered square $101 \times 101$, each cell of the inner square $99 \times 99$ is painted in one of ten colors (cells adjacent to the border of the square, not painted). Could it turn out that in every in a $3\times 3$ square, is exactly one more cell painted the same color as the central cell?
Let $n > 1$ be an odd natural number. The squares of an $n \times n$ chessboard are alternately colored white and black so that the four corner squares are black. An $L$-triomino is an $L$-shaped piece that covers exactly three squares of the board. For which values of $n$ is it possible to cover all black squares with $L$-triominoes, so that no two $L$-triominos overlap? For these values of $n$ determine the smallest possible number of $L$-triominoes that are necessary for this.
Find the largest natural number $n$ such that any set of $n$ tetraminoes, each of which is one of the four shapes in the picture, can be placed without overlapping in a $20 \times 20$ table (no tetramino extends beyond the borders of the table), such that each tetramino covers exactly 4 cells of the 20x20 table. An individual tetramino is allowed to turn and flip at will.
[img]https://cdn.artofproblemsolving.com/attachments/b/9/0dddb25c2aa07536b711ded8363679e47972d6.png[/img]
With $21$ tiles, some white and some black, a $3 \times 7$ rectangle is formed. Show that there are always four tokens of the same color located at the vertices of a rectangle.