Subfield: Number Theory


Question: An L-shaped triomino is composed of three unit squares, as shown:

![](https://cdn.mathpix.com/cropped/2023_12_21_96cde5457e2677371bf4g-1.jpg?height=130&width=146&top_left_y=1553&top_left_x=1518)

Suppose that $H$ and $W$ are positive integers. An $H \times W$ rectangle can be tiled if the rectangle can be completely covered with non-overlapping copies of this triomino (each of which can be rotated and/or translated) and the sum of the areas of these non-overlapping triominos equals the area of the rectangle (that is, no triomino is partly outside the rectangle). If such a rectangle can be tiled, a tiling is a specific configuration of triominos that tile the rectangle.
Determine, with justification, all pairs $(H, W)$ of integers with $H \geq 4$ and $W \geq 4$ for which an $H \times W$ rectangle can be tiled.

Solution:
Suppose that $(H, W)$ is a pair of integers with $H \geq 4$ and $W \geq 4$.

Since the area of each triomino is 3 , then the area of any rectangle that can be tiled must be a multiple of 3 since it is completely covered by triominos with area 3 .

Since the area of an $H \times W$ rectangle is $H W$, then we need $H W$ to be a multiple of 3 , which means that at least one of $H$ and $W$ is a multiple of 3.

Since a rectangle that is $a \times b$ can be tiled if and only if a rectangle that is $b \times a$ can be tiled (we see this by rotating the tilings by $90^{\circ}$ as we did with the $3 \times 2$ and $2 \times 3$ rectangles above), then we may assume without loss of generality that $H$ is divisible by 3 .

We show that if $H$ is divisible by 3 , then every $H \times W$ rectangle with $H \geq 4$ and $W \geq 4$ can be tiled.

Case 1: $H$ is divisible by $3, H$ is even

Here, $H$ is a multiple of 6 , say $H=6 m$ for some positive integer $m$.

Since $W \geq 4$, we know that a $6 \times W$ rectangle can be tiled.

By stacking $m 6 \times W$ rectangles on top of each other, we obtain a $6 m \times W$ rectangle.

Since each $6 \times W$ rectangle can be tiled, then the $6 m \times W$ rectangle can be tiled.

Case 2: $H$ is divisible by $3, H$ is odd, $W$ is even

Suppose that $H=3 q$ for some odd positive integer $q$ and $W=2 r$ for some positive integer $r$.

To tile a $3 q \times 2 r$ rectangle, we combine $q r 3 \times 2$ rectangles in $q$ rows and $r$ columns:

![](https://cdn.mathpix.com/cropped/2023_12_21_742b0b713146d818e937g-1.jpg?height=258&width=219&top_left_y=1324&top_left_x=1056)

Therefore, every rectangle in this case can be tiled. (Note that in this case the fact that $q$ was odd was not important.)

Case 3: $H$ is divisible by $3, H$ is odd, $W$ is odd

Since $H \geq 4$ and $W \geq 4$, the rectangle with the smallest values of $H$ and $W$ is $9 \times 5$ which can be tiled as shown:

![](https://cdn.mathpix.com/cropped/2023_12_21_742b0b713146d818e937g-1.jpg?height=559&width=330&top_left_y=1886&top_left_x=995)

(There are also other ways to tile this rectangle.)

Since $H$ is an odd multiple of 3 and $H \geq 4$, we can write $H=9+6 s$ for some integer $s \geq 0$.



Since $W$ is odd and $W \geq 5$, we can write $W=5+2 t$ for some integer $t \geq 0$.

Thus, the $H \times W$ rectangle is $(9+6 s) \times(5+2 t)$.

We break this rectangle into three rectangles - one that is $9 \times 5$, one that is $9 \times 2 t$, and one that is $6 s \times W$ 

(If $s=0$ or $t=0$, there will be fewer than three rectangles.)

The $9 \times 5$ rectangle can be tiled as we showed earlier.

If $t>0$, the $9 \times 2 t$ rectangle can be tiled as seen in Case 2 .

If $s>0$, the $6 s \times W$ rectangle can be tiled as seen in Case 1 .

Therefore, the $H \times W$ rectangle can be tiled.

Through these three cases, we have shown that any $H \times W$ rectangle with $H \geq 4$ and $W \geq 4$ can be tiled when $H$ is a multiple of 3 .

Since the roles of $H$ and $W$ can be interchanged and since at least one of $H$ and $W$ must be a multiple of 3 , then an $H \times W$ rectangle with $H \geq 4$ and $W \geq 4$ can be tiled exactly when at least one of $H$ and $W$ is a multiple of 3 .

Final answer: exactly when at least one of $H$ and $W$ is a multiple of 3