In the beginnig, all nine squares of $3\times 3$ chessboard contain $0$. At each step, we choose two squares sharing a common edge, then we add $1$ to them or $-1$ to them. Show that it is not possible to make all squares $2$, after a finite number of steps.
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tenniskidperson3
20.01.2013 23:36
The sum of the elements in the corners and middle are equal to the sum of the elements in the edge squares. Not true when all squares are 2.
94337sk
30.03.2015 09:02
also a change is occuring in sets of 2 which is an even no. but there are odd no. of boxes.
IstekOlympiadTeam
08.04.2015 16:01
I solved with $ mod3$
BarisKoyuncu
25.07.2019 16:57
We can paint chessboard with black and white. In the beginning sum of black squares equal sum of white squares and each step affects one black and one white square. Then sum of black squares equal sum of white squares every time. Then all squares can not be 2.
mathstudent5
03.02.2024 18:31
The sum is invariant . In every step, the sum is 0. But the sum needs to be 18. A contradiction