[Guest]

This is pretty simple:

The normal chess board can be covered entirely using 32 rectangular cardboards with each cardboard covering exactly 2 squares.

Now, if you remove the bottom left and top right square of the chess board, is it still coverable with 31 rectangular cardboards?

If yes, then show how else prove the impossibility.

[grey cells]

I think not .

Because as the 2 end columns contain odd no. of squares .

Just first thoughts .

[Guest]

left is sq 1 -( blank ) then cover, 2n3 4n5 6n7 8n8b B being next column, 7b+6b and continue cornering accordingly.

I suppose the reasoning that the missing squares are at opposing ends so it would not be a jump and that there are an even number of squares

[Guest]

I think not .

Because as the 2 end columns contain odd no. of squares .

Just first thoughts .

Good guess, but even if the columns or rows contain odd number of squares, it would still be coverable by the rectangles.

If you say no, you must provide a valid proof.

[Guest]

left is sq 1 -( blank ) then cover, 2n3 4n5 6n7 8n8b B being next column, 7b+6b and continue cornering accordingly.

I suppose the reasoning that the missing squares are at opposing ends so it would not be a jump and that there are an even number of squares

Good reasoning

But can you extend it to the whole board?

If yes then please show.

[Guest]

This is pretty simple:

The normal chess board can be covered entirely using 32 rectangular cardboards with each cardboard covering exactly 2 squares.

Now, if you remove the bottom left and top right square of the chess board, is it still coverable with 31 rectangular cardboards?

If yes, then show how else prove the impossibility.

I think it is impossible. I cannot say this is a proof, but the idea is that each rectangle cardboard covers one white and one black squares. Since we have removed squares of the same color, it might be impossible to cover all the squares with the rectangular cardboards.

[Guest]

Exactly brhan, you got it. Didn't I say, its simple.

A rectangle always covers 2 different colours. The 2 squares which we have removed from the board are of same colour, therefore the the board has 2 more squares of one colour than the other. And after the rectangles have covered the 60 squares, there will left 2 squares of the same colour. And obviously, the remaining rectangle would not cover those 2 squares.