Let me go grab a hamburger real quick...

Ok, I'm back.

How many **squares** are there in the $6\times4$ grid below?
That's a reaallly good question!

Let's start by counting the smallest $1\times1$ squares, this is just the same as counting the number of unit squares in a $6\times4$ grid, there are $6\times4=24$ 1 by 1 squares in the grid.

Now let's move on to the $2\times2$ squares, notice that counting the number of $2\times2$ squares in a $6\times4$ grid is just the same as counting the number of unit squares in a $5\times3$ grid. So the number of 2 by 2 squares in the grid is $5\times3=15$.

Now the $3\times3$ squares, similarly, counting the number of $3\times3$ squares in a $6\times4$ grid is just the same as counting the number of unit squares in a $4\times2$ grid, which is $4\times2=8$.

And again the number of $4\times4$ squares in a $6\times4$ grid is equal to the number of unit squares in a $3\times1$ grid, which is $3\times1=3$.

Add up all the number of squares together: $24+15+8+3=50$. Tada! We now have our answer! There are 50 squares in a $6\times4$ grid.

Mmm... the hamburger is really good...

Back on topic, in general, what is the total number of squares in an $a\times b$ grid (where $a$ is the **width** of the grid and $b$ is the **height** of the grid), given $a\geqslant b$?

Again let's start from the $1\times1$ squares, that's trivial, there's $ab$ of them.

Now moving on to the $2\times2$ squares, the number of $2\times2$ squares in an $a\times b$ grid is equal to the number of unit squares in an $(a-1)\times(b-1)$ grid.

Notice the pattern? Counting the number of $n\times n$ squares in an $a\times b$ grid is the same as counting the number of unit squares in an $(a-n+1)\times(b-n+1)$ grid.

The largest square that can contain in an $a\times b$ grid given that $a\geqslant b$ is $b\times b$.

Hence, the total number of squares in an $a\times b$ grid is $ab+(a-1)(b-1)+(a-2)(b-2)+\ldots+[a-(b-2)][b-(b-2)]+[a-(b-1)][b-(b-1)]$ Or $\sum_{i=0}^{b-1}{(a-i)(b-i)}$

This is ugly, we don't like sigma symbols sitting around, so why not we simplify this a little bit...

$\begin{aligned} \sum_{i=0}^{b-1}{(a-i)(b-i)}&=\sum_{i=0}^{b-1}{[ab-(a+b)i+i^2]} \\&=ab^2-\frac{(a+b)b(b-1)}{2}+\frac{b(b-1)(2b-1)}{6} \\&=b\left[ab-\frac{ab-a+b^2-b}{2}+\frac{2b^2-3b+1}{6}\right] \\&=\frac{b}{6}\left[6ab-3ab+3a-3b^2+3b+2b^2-3b+1\right] \\&=\frac{b}{6}\left[3ab+3a-b^2+1\right] \\&=\frac{b(b+1)(3a-b+1)}{6} \end{aligned}$ BOOM! There we have it! *Round of applause* *Fireworks* *Pancakes*

The total number of squares in an $a\times b$ grid (where $a$ is the

widthof the grid and $b$ is theheightof the grid), given $a\geqslant b$ is $\frac{b(b+1)(3a-b+1)}{6}$If $a<b$, then we just swap $a$ and $b$.

Special case:If $a=b$, the above equation becomes $\frac{a(a+1)(2a+1)}{6}$ which is the formula for the sum of squares from 1 to $a$.

Done! Now let me finish my burger...

No vote yet

1 vote

Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in`\(`

...`\)`

or`\[`

...`\]`

to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestNice simple way of explaining complex situation. So many thanks.

Log in to reply

Cool! Thanks for that note bro, you're awesome!

Log in to reply

Hope you finished your burger peacefully :P

Log in to reply

Thanks, I'm glad you liked the note. Well unfortunately, I think my hamburger has become stale. XD

Log in to reply

You should add this to the Brilliant wiki. Great note!

Log in to reply

are you a robot, cos I need some real friends? Humanity is a lie, the computer generation is upon us. Support the cause m64^(1/2)

Log in to reply

No, i am 100.1% sure I'm not a robot.

Log in to reply

What a note @Tan Kenneth:)

Log in to reply

HEY tankenneth, you hyped?

Log in to reply

Oh yes I am! $1+1=3$

Log in to reply

جميلة

Log in to reply

Translation: beautiful!

Log in to reply