A nice easy, sort of, chess puzzle.

As a maths teacher, I set this problem for my Year 7s today and gave them three minutes to find the solution and provide a little mathematical arguement. Can you do the same?

How many squares are there on a chess board?

The answer is not 64!

204.

64 little squares,16 2x2 squares,4 4x4 squares,4 3x3 squares,1 5x5,1 6x6,1 7x7,1 8x8, so i am going to say 92 squares.

sniperghost360 wrote:

64 little squares,16 2x2 squares,4 4x4 squares,4 3x3 squares,1 5x5,1 6x6,1 7x7,1 8x8, so i am going to say 92 squares.

Actually, there are 64 1x1 squares, 49 2x2 squares (count them), 36 3x3 squares (count them), 25 4x4 squares (count them), 16 5x5 squares (count them), 9 6x6 squares (count them), 4 7x7 squares (count them), and 1 8x8 square.

64+49+36+25+16+9+4+1 = 204 squares, so reflaxion got there before me.

206?

reflaxion and dmeng get one merit and move to the top of the class.

The answer is a summation of the square numbers from 1^2 to 8^2 which equals 204.

cool..wish i was good at math

There are only two squares on the board --- the front and back .The area to move is just a logistists  of the game.

To get back on topic

Always try to defend your pieces more than they are attacked. If it's attacked once try to defend it twice, if it's attacked twice try to defend it 3 times and so on.

Please ignore

Posted in wrong topic

Now from the diagram shown (White king at a1 and Black king at a8), for sure white king could have captured Black king after 7 consecutive moves. But how many ways are there to make such series of moves?

NQChien, if you mean capture the black king in only 7 consecutive moves, I would say there's only 1 way.

With the Kings on a1 and h8, yes.

But on a1 and a8, a few more, but not as many as on e1 and e8.

 Yes and if that fails, move the piece or attack opponent's unguarded piece of equal or greater value.

1296. I can post the justification for that if you want. :)

Don't keep us in suspenders!!

No doubt the proof is too large for the margin to contain.

I'll take that as a "Yes, go ahead and post it."

Drawing any 8x8 grid like that takes 9 horizontal lines and 9 vertical lines. Each rectangle can be made by picking two horizontal lines (for the top and bottom sides), and two vertical lines (for the left and right sides).

So, there are 9!/(2!7!) = 36 ways to choose the top and bottom, as well as 9!/(2!7!) = 36 ways to choose the left and right. This makes 36*36 = 1296 ways to choose all four sides of the rectangle, and that is your total number.

@ tyzebug - No, but it's an amusing reference anyway.

Tut tut. It's 1968.

The formula is (x^2)[(9-x)^2)], 0<9 where x is an integer and is the length of the sides of the square.

There are many lengths including;

1x1, 2x2, 3x3, ... , 8x8

You must add all the possibilities.

(1^2)[(9-1)^2)]+(2^2)[(9-2)^2)]+(3^2)[(9-3)^2)]+...+(8^2)[(9-8)^2)]

There,

1968 if I'm not mistaken.