anonymous
  • anonymous
Assume an \(8\times 8\) chessboard with usual colouring. You may repaint all the squares of a row or column. The goal is to attain one black square. Possible?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
yes
anonymous
  • anonymous
i think
anonymous
  • anonymous
do u need explanation

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More answers

terenzreignz
  • terenzreignz
What does repaint mean? Invert the colours of a row? (IE turn all light squares dark and vice versa)
anonymous
  • anonymous
ok suppose that we want one black square that is how we can do it
anonymous
  • anonymous
black square is in the middle lets say at 4rth row and 4rth coloumn
Jack1
  • Jack1
yes
anonymous
  • anonymous
first we paint all the rows except 4rth one with white color not 4rth row 4rth row must have one black
anonymous
  • anonymous
then we paint all coloumns with white except 4rth coloumn
terenzreignz
  • terenzreignz
So, in simple terms, repainting is a function that maps... |dw:1370257845631:dw|
anonymous
  • anonymous
in this way all the remaining except the the one we said will be white and only one which is at location 4rth row and 4rth coloumn would be white
anonymous
  • anonymous
so ans is yes
anonymous
  • anonymous
ok repainting means painting what you want to paint if 1 row is to be repainted with white then all the squares in that row would be white
anonymous
  • anonymous
I guess so @terenzreignz , because the answer at back is 'no'.
terenzreignz
  • terenzreignz
Well then, \[\large \left[\begin{matrix}0&1&0&1&0&1&0&1 \\1&0&1&0&1&0&1&0\\0&1&0&1&0&1&0&1\\1&0&1&0&1&0&1&0\\0&1&0&1&0&1&0&1\\1&0&1&0&1&0&1&0\\0&1&0&1&0&1&0&1\\1&0&1&0&1&0&1&0\end{matrix}\right]\] Okay, let's start with this... 0 = light square 1 = dark square
terenzreignz
  • terenzreignz
Let's repaint the 1st, 3rd, 5th and 7th rows (starting from the top) 1&0&1&0&1&0&1&0 \[\large \left[\begin{matrix}1&0&1&0&1&0&1&0\\1&0&1&0&1&0&1&0\\1&0&1&0&1&0&1&0\\1&0&1&0&1&0&1&0\\1&0&1&0&1&0&1&0\\1&0&1&0&1&0&1&0\\1&0&1&0&1&0&1&0\\1&0&1&0&1&0&1&0\end{matrix}\right]\]
terenzreignz
  • terenzreignz
Okay, nvm, I lost myself :D
terenzreignz
  • terenzreignz
It might the case that no matter how you repaint, you always add/subtract an even number of dark squares.... so you can't end up with just 1...
terenzreignz
  • terenzreignz
Though you could always repaint the second, fourth, sixth, and eight columns (from the left) and end up with one (really big) dark square XD
anonymous
  • anonymous
That's it.

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