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In a square of dimensions $100 \times 100$ each $1 \times 1$ square is white or black. We are permitted to change the colors of all $1 \times 1$ squares on the same line or on the same column. If initially all $100 \times 100$ squares are white, is it possible to construct a succession of transformations that results in exactly $760$ black squares?

N. F. Taussig
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Adele
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2 Answers2

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It isn’t possible. If we toggle $m$ rows and $n$ columns, there are $m(100-n)+(100-m)n$ black squares, so we’d need to have $m(100-n)+(100-m)n=760$. Solving for $n$ yields

$$ n=\frac{760-100m}{100-2m}\;. $$

This is negative for $8\le m\le49$, undefined for $m=50$ and greater than $100$ for $51\le m\le92$. Thus you just need to check that it’s not an integer for $0\le m\le7$ and $93\le m\le100$; see here and here.

joriki
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  • Thank you but how can you prove that there are m(100-n)+(100-m)n black squares? – Adele Feb 01 '20 at 18:03
  • @Adele: Is that still an open question or has it been resolved through your exchange with Rob Pratt under the other answer? – joriki Feb 01 '20 at 21:25
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Hints: Without loss of generality, each row or column is chosen at most once, because choosing the same row or column twice has no effect. Also without loss of generality, only the first $r$ rows and first $c$ columns are chosen, because permuting rows and columns does not change the number of black squares. Now the top left corner consists of $rc$ white squares, the bottom right corner consists of $(100-r)(100-c)$ white squares, and the rest are black. This yields $r(100-c)+c(100-r)$ black squares, as in @joriki’s answer.

RobPratt
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