Possible Checker Placements

Number the squares of the grid with the numbers $1,2,3$ in the following pattern: $\begin{array}{|c|c|c|c|c|c|c|c|c|c} \hline 1 & 2 & 3 & 1 & 2 & 3 & 1 & 2 & 3 & \cdots \\ \hline 2 & 3 & 1 & 2 & 3 & 1 & 2 & 3 & 1 & \cdots \\ \hline 3 & 1 & 2 & 3 & 1 & 2 & 3 & 1 & 2 & \cdots \\ \hline 1 & 2 & 3 & 1 & 2 & 3 & 1 & 2 & 3 & \cdots \\ \hline 2 & 3 & 1 & 2 & 3 & 1 & 2 & 3 & 1 & \cdots \\ \hline 3 & 1 & 2 & 3 & 1 & 2 & 3 & 1 & 2 & \cdots \\ \hline 1 & 2 & 3 & 1 & 2 & 3 & 1 & 2 & 3 & \cdots \\ \hline 2 & 3 & 1 & 2 & 3 & 1 & 2 & 3 & 1 & \cdots \\ \hline 3 & 1 & 2 & 3 & 1 & 2 & 3 & 1 & 2 & \cdots \\ \hline \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots \\ \end{array}$ There are $534$ squares labelled $1$ and $533$ labelled $2$ and $3$ each. Thus the checker piece must be placed on a square labelled $1$ , since every trimonoe covers one square labelled $1$ , one square labelled $2$ and one square labelled $3$ .

However, if the board is rotated through $90^\circ$ , then the squares originally labelled $1$ in rows $2,3,5,6,8,9,11,12,\ldots$ are now labelled $2$ or $3$ - the only squares that are labelled with $1$ in both the original and the rotated position are the squares labelled $1$ in rows $1,4,7,10,\ldots,40$ . There are $14$ such squares in each of these $14$ rows, and hence there are $196$ squares that interest us.

It is each to show that a tiling with trimonoes is possible if the checker piece is placed on any of these $196$ squares: the remaining squares in the same row as the checker piece can be tiled with trimonoes, and the remaining $39$ rows come neatly in groups of three, and each group of three can readily be tiled by $40$ vertically oriented trimonoes. Thus the answer is indeed $196$ .

Label the x and y coordinates from 0 to 39. We claim that the valid checker positions are precisely those with $x \equiv y \equiv 0 \pmod 3$ . First we show that the condition is necessary: 3-colour the squares of the grid in a checkerboard pattern according to the value of $x + y \pmod 3$ . Then of the 1600 squares, exactly 534 will be coloured 0, while colours 1 and 2 occur 533 times each. Since any triomino covers exactly one square of each colour, the checker must lie on a square coloured 0: if the coordinates are $(x_0,y_0)$ then $x_0+y_0 \equiv 0 \pmod 3$ . An identical argument with the colouring $x - y \pmod 3$ shows that also $x_0-y_0 \equiv 0 \pmod 3$ , which gives $x_0 \equiv y_0 \equiv 0 \pmod 3$ as claimed.

Conversely, if we place the checker at a position with both x and y divisible by 3, it's easy to see that the row it is in can be tiled with triominos, and that the subrectangles above and below that row each have height divisible by 3, and thus can be likewise tiled. Thus the condition is also sufficient, and the number of squares satisfying it is $14^2 = 196$ ( $x/3$ lies between 0 and 13 so there are 14 choices for x and similarly for y).

Assign point values of $4, -2,$ and $1$ to the squares according to this pattern:

$4, -2, -2, ...$

$-2, 1, 1,...$

$-2, 1, 1,...$

where each row or column keeps repeating the same three numbers (either $4, -2, -2$ or $-2, 1, 1$ ).

If the checker piece is on a square with value $4$ then the rest can be covered with triminos. (For example, first cover the rest of the checker's row and column. This leaves four (or fewer) rectangles whose dimensions are all multiples of three, which can obviously be covered with triminos.)

But on the other hand, the triminos must cover squares with point-value total of exactly zero (each trimino covers either $4, -2, -2$ or $-2, 1, 1$ ), so all possible locations of the checker must have the same point value.

And since we already showed that every square with point value $4$ is a possible location for the checker piece, the answer will be the number of such squares or $14^2=196$ .

As the values ofK's are increasing, we take the grids by trimming the previous grid along all the 4 boundaries. So the required answer is .

We can cover any $3 \times 40$ rectangle by filling first the $3 \times 39$ part, and then filling the last $1 \times 3$ part with a vertical tetromino (or horizontal, if you work from side to side). So we can always fill the $39 \times 40$ part of the board. For the row or column we have at this moment, we put $13$ tetrominos, and we can move the tetrominos from $0$ to $13$ times. This means we have in that row $13 - 0 + 1 = 14$ spaces for the checkers piece. But this can't be repeated in all the rows. We can repeat it whenever we can move one or more $3 \times 40$ rectangle (because we cant break the tetrominos). So again we have $13$ of those rectangles, which gives $14$ rows in which we can make the $14$ other operations. Important: Don't repeat this from side to side if you did it vertically, because you'll repeat the same spaces. So the answer is $14 \cdot 14 = \boxed {196}$ .

You can think the square grid as a collection of cells with 0 1 or 2 written on it. 0 1 2 0 1 2 0 ...... 1 2 0 1 2 0 1 ...... 2 0 1 2 0 1 2 ....... ........................

Now each trimino can be placed on 012 thus the total 3

As the sum of all elements is divisible by 3, so the covered cell must be a 0 cell. There are 14*14=196 0 cells. so ans is 196

Colour the diagonals that go in the bottom-left to top-right direction in three alternating colours: red, white, and blue. Then the diagonals that are coloured red will be of lengths $1, 4, 7, 10, ..., 37, 40, 37, ..., 4, 1.$ So the number of red squares is $40 + 2\times \sum_{n=1}^{12} (1+3n) = 40 + 2(12 + 3\times \frac{13 \times 12}{2}) = 532.$ The diagonals that are coloured white will be of lengths $2, 5, 8, ..., 38, 39, 36, ..., 6, 3.$ So the number of white squares is $\sum_{n=1}^{12} (2+3n) + \sum_{n=1}^{12} (3+3n) = 351.$ The diagonals that are coloured blue will be of lengths $3, 6, 9, ..., 39, 38, 35, ..., 5, 2,$ so that the number of blue square is also 351.

Since with this colouring of the squares, any trimino covers a red, a white, and a blue square, a tiling by triminos will colour the same number of red, white, and blue squares. Therefore, Victor must place the checker piece on a red square.

Furthermore, if we colour the squares down the diagonals in the other direction so that the diagonals go in the top-left to bottom-right direction, we can apply the same argument. Then Victor can only place the checker piece on squares that will be coloured red in both colouring schemes. These squares form a $14 \times 14$ grid, with a total of 196 squares.

Conversely, we can always tile the grid after Victor covers one of these 196 tiles: it is easy to see that one can tile a $4 \times 4$ square grid with triminos if one of the corners is covered. When Victor places a checker piece over a square, draw a $4 \times 4$ grid around it so that the covered square is in the corner. Then we can tile that $4 \times 4$ grid, and we can also tile the remaining board because it can be broken up into rectangles where each side is a multiple of three squares wide.

For $(40-3*2*k) \times (40-3*2*k)$ grid we find $(40-1-3*2*k)/3 \times (40-1-3*2*k)/3 = 4*(40-1-3*2*k)/3$ different places to put the checker piece and the positions are $3*i+1$ for first and last row where $i = 0, 1, ..., (40-1-3*2*k)/3$ and $3*j+1$ for first and last column where $j = 1, 2, ..., (40-1-3*2*k)/3-1$ and $k = 0, 1, ..., 6$ .