1. A T-tetromino is formed by adjoining three unit squares to form a $1 \times 3$ rectangle, and adjoining on top of the middle square a fourth unit square. Determine the minimum number of unit squares that must be removed from a $202 \times 202$ grid so that it can be tiled with T-tetrominoes.
Solution. We provide the following construction, which shows that the answer is at most four. Clearly, four T-tetrominoes tile a $4 \times 4$ grid, as follows:

These can be used to tile the upper-left $200 \times 200$ grid. It can be verified that the following pattern fills the remaining space, leaving four grid squares uncovered:

As the answer must be a multiple of four, it remains to prove that the answer is not zero, that is, it is impossible to tile a $202 \times 202$ grid with T-tetrominoes.

Number the rows of the grid, from left to right, with $0,1, \ldots, 201$. Similarly, number the columns of the grid, from top to bottom, with $0,1, \ldots, 201$.
Assign the square numbered $(x, y)$ with the weight $4 x+4 y+1$. It can be verified that a T-tetromino, no matter how it is placed, covers a sum of weights divisible by 8 . However, the sum of all the weights is not divisible by 8 , as desired.

Solution 2: An alternative way to prove that the $202 \times 202$ grid cannot be tiled with T-tetrominoes is with a checkerboard coloring.
For the sake of contradiction, assume that the board can be tiled with T-tetrominoes. Clearly, a T-tetromino covers only either three or one black square. Let $x$ be the number of T-tetrominoes each covering three black squares and $y$ be the number of T-tetrominoes each covering only one black square. Counting the number of black and white squares yields
$$
3 x+y=2 \cdot 101^{2}, \quad x+3 y=2 \cdot 101^{2} \quad \Longrightarrow \quad x=y=\frac{101^{2}}{2} .
$$
which contradicts the fact that $x$ and $y$ are integers.