Squares in rectangles poster

There are three rectangles which contain $100$ squares:  $1 \times 100$, $4 \times 11$ and $5 \times 8$

If the height is $1$, increasing the width by $1$ increases the number of squares by $1$. This is because the only size of square that we can make is $1 \times 1$, so adding $1$ to the width adds just $1$ square. Eventually, when the rectangle is $1 \times 100$ there will be $100$ rectangles.

If the height is $2$, increasing the width by $1$ increases the number of squares by $3$. This is because when we add $1$ to the width, we can make $2$ additional $1 \times 1$ squares and $1$ additional $2 \times 2$ square. This is a total of $3$ extra squares.

If we continue the pattern $2, 5, 8, \ldots$ in the height of $2$ row (all $1$ less than multiples of $3$), we eventually get to $\ldots, 98, 101, \ldots$ - missing out $100$. This tells us that it isn't possible to make a rectangle with $100$ squares when the height (or width) is $2$.

If the height is $3$, increasing the width by $1$ allows us to make $3$ more $1 \times 1$ squares, $2$ more $2 \times 2$ squares and $1$ more $3 \times 3$ square. This is a total of $6 = 3 + 2 + 1$ squares. This gives us the sequence $14, 20, 26, 32, \ldots, 98, 104, \ldots$ (all $2$ more than multiples of $6$) which tells us that we can't make a rectangle with exactly $100$ squares when the height (or width) is $3$.

Using the same reasoning for height of $4$, we see that increasing the width by $1$ increases the number of squares by $10$, and that we can make rectangles with $20, 30, \ldots, 90, 100, 110, \ldots$ squares. $100$ is in this list! In fact, a $4 \times 11$ rectangle contains exactly $100$ squares.

We can repeat this for the other heights (or widths) in the table to find all the rectangles with exactly $100$ squares. These are $1 \times 100$, $4 \times 11$ and $5 \times 8$.