On Friday I set this puzzle….. how many squares are there on an 8×8 chessboard?

If you have not tried to solve it, have a go now. For everyone else, the answer is after the break.

204! Did you solve it?

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Yeap: you forgot the square on the back of the chessboard 🙂 That makes a real chessboard have 205

(and even more if it has a border 😉 ).

Could you show your workings please. I’d like the opportunity to learn something.

Guess I can Google it.

smallest squares: 8 x 8 on a board

squares of 2 x 2 black/white combi: 7 x7 on a board

etc, making it

8² + 7² + 6² + … + 1 = 204

+ the one on the back = 205

Ignoring the one on the back, what Roland said is the method I used.

And really, if your chessboard is three-dimensional, and if you are going to be as technical as some people here, perhaps you should count the outside rim as one, also…

so, 206! 😀

I didn’t reference an actual board, but my rough mind’s eye calculation came to just under 200. I was on the right path.

Yes, got it.

8^2+7^2+6^2+5^2+4^2+3^2+2^2+1^2=

64+49+36+25+16+9+4+1=

204

That’s all. Next week: how many rectangles are there on a chess board?

That would be pure torture :))

As I posted on the puzzle-site from Friday:

For Rectangles I found the following formula

Sum of n³ for n from 1 to 8 = 8² * 9² / 4 = 1296

yup, 204 squares, 1296 rectangles, how many arrangments of chessmen?

I likr that

Got it! After taking pencil and paper and outlining just a few of the different size squares and their possible locations, it was easy to see the pattern.

For example, there are 8 possible locations across and 8 possible locations down for a square of 1 unit, making 8×8=64.

Next, there are 7 possible locations across and 7 possible locations down for a square of 4 units, making 7×7=49.

This goes on until, the last one being 1 possible location across and 1 possible locatin down for a square of 64 units, making 1×1=1.

So 64+49+36+25+16+9+4+1=204

Of course, this generalises to any n*n chess board:

(sum of i^2 for i from 1 to n) = n (n + 1) (2n +1) / 6

So for n = 8, we get 8*9*17/6 = 204

(See http://www.wolframalpha.com/input/?i=sum+i^2+for+i+from+1+to+n)

Yes, thanks for posting the formula.

Since the sum i for i from 1 to n = n(n+1)/2, it can be proven that the sum of i^2 for i from 1 to n = n^3/3 + n^2/2 + n/6

( n = 8, 170.67+32+1.33 = 204 )

which can be simplified to n(n+1)(2n+1)/6 for easier substitution as you show. Thank You!

Yesss!

Little script (using ‘recursion’) answers for any size board: http://jsfiddle.net/v5tWG/1/

Easy puzzle this week 🙂 Nothing out of the box required really.

But hey, actually the real answer is infinity.

You can fit an infinite number of squares on any board or square. The chess board is 8×8 units big, so surely a whole lot of squares the size of 0.001×0.001, 0.002×0.002, 0.003×0.003 etc would fit.

hmm rectangles… lets see, by definition, squares are also rectangles.

So, that would be 204 + 8*7*2 + 8*6*2 + 8*5*2 + 8*4*2 + 8*3*2 + 8*2*2 + 8*1*2 + 7*6*2+ 7*5*2 + 7*4*3 +……….

the 8*x*2 series is for y*1 vertical rectangle + 2*1 horizontal rectangles. 7*x*2 is for y*2 rectangle, then you need to do y*4, y*5 etc.

OK, you tricked me. You’d think I’d expect it by now, but yes, I said 64. And I’m OK with that.

It said nothing about overlapping squares. You can have thousands of those

204 for me…I lost the one in the back of the chessboard, XD.

I used this Perl one-line program to calculate the 204:

perl -MList::Util=sum -E ‘say sum map { $_ * $_ } 1 .. 8’

well i got 204, it was quite straight forward. doing

how many square that are:

8×8

7×7

6×6

5×5

4×4

3×3

2×2

1×1

this is what our maths class did 🙂

this is a batty man puzle.

KMT. it was too hard bruv.

Y U GONNA COME GIVE ME SOME NE NXT answer?

ur a true numba jack.

this aint even in english lyk. there is no mention of peng ne where.

SNM. (in maths language QED.)

This is pretty much how all British people sound in my head.

🙂

I got 204, but unlike most posters above, I started with the one big square, then I visualized the 4 7×7 squares. When I got to the 9 6×6 squares I noticed that so far I had results of 1, 4 and 9 which I always remember as the dimensions of the monolith in 2001 A Space Oddesey and the first three perfect squares. I was kinda jazzed to see the pattern continue up to the 64 small squares. So anyway, I guess I did it backwards!

No, I didn’t get it because I tried to just brute-force count them and lost the energy to care around 96. I’m smart enough to know the question intends that we include squares of different sizes and not just the 64 you can move on, but I’m not smart enough to write a formula to do it.

This is another excellent mind boggling puzzle. Thanks again richard! i look forward to the next installment on friday. I appreciate your work son!

hahahaha looking forward to meeting up for lunch one day.

http://www.standupmaths.com/

this is my website feel free to check it out.

I got the idea, but failed to see the pattern. Of the explanations above, I found Jamie’s the easiest to follow.

The correct answer is obviously 64.

If I showed you a pattern of two white squares and two black ones, no one in his senses would call that pattern “a square”.

204 is the correct anser two the question “How many different, but overlapping squares could you draw along the borders of the existing squares on a chess board?”

I have only just looked at this puzzle and got the wrong answer. I then read the comments and realised wherer I was going wrong. I wonder if I would have realised on my own had a picture of a chess board been shown?

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