Interesting Math/Reasoning problem (check my work)

[Scrapster]

Problem:
Two squares from each of two opposite corners are deleted from an eight by eight checkerboard. Prove that the remaining squares cannot be covered exactly by copies of the "T-Shape" and its rotations. (A "T-Shape" consists of 4 squares).

My answer:
If we remove 2 squares from each of opposite corners we are left with 60 squares left (64-4=60): 30 white and 30 black (We'll say the checkerboard is B&W). A "T" takes up 4 sqaures so in order to completely cover the board we would need 15 "T's" (60/4=15). A "T" can consist of 3 white and 1 black (3w:1b) or 3 black and 1 white (1w:3b).

I've tried a few different arrangements and the most T's I can come up with is 14, that's with the "T" arrangements being half and half (7 1w:3b and 7 3w:1b). I am positive that this is the maximum T's we can come up with (though I haven't manually calculated EVERY POSSIBLE arrangement), but the few other scenarios I've tried keep giving me less than 14. So my proof is pretty much complete, but I want to make it "airtight". Do you guys see any opportunity for better reasoning?

 

[Scrapster]

Any input?

 

[Scrapster]

Anyone? Please.

 

[dopcombo]

er, i dun see how a T takes up only 4 squares

 

[dopcombo]

and if that's the ORiginal qn, i think they mean, remove only 2 squares to give 62 squares left....

 

[Scrapster]

The problem was directly out of the book. and a T is made up of 4 squares.

o
oo <---T
o

Please check my work guys.

 

[thEnEuRoMancER]

In order to cover the remaining part of the checkerboard we need to place 15 pieces of &quot;T&quot; on it. Some of them (x) will be (3w:1b) and some of them (15-x) will be (1w:3b). Of the remaining 60 squares, 30 are black and 30 are white and we can write the following equation (basically counting the white and black squares:

x(3w+1b) + (15-x)(1w+3b) = 30w + 30b

solve this and obtain:

w(2x+15) + b(45-2x) = 30w + 30b

2x+15=30
45-2x=30

x must be an integer, but the only solution is x=7.5 . This proves it is impossible to cover the checkerboard with x pieces of 3w:1b 'T' and (15-x) pieces of 1w:3b 'T', where x is an integer.

BTW: I can't see the proof in your reasoning, Scrapster. There are probably literally millions of different combinations and there's no way you have tried nearly all of them.

 

[thEnEuRoMancER]

BTW: this is yet another B&amp;W thread