Each cell of a 1997x1997 square grid contains either +1 or 1, with no cell being vacant.
The product of all the numbers in the ith row, and
the product of all the numbers in the ith column are respectively denoted by R_{i} and C_{i}.
1997
Prove that Σ_{i=1}(R_{i} + C_{i}) is always nonzero.
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I think this should do it. Consider any grid with X rows and Y columns
where all cells contain the value 1. Σ_{i=1}(R_{i} + C_{i})
= X + Y.
<o:p> </o:p>
From here we have two possible moves:
<o:p> </o:p>
(i)
If we change the sign of cell
(X_{i} , Y_{i} ) where R_{i} and C_{i} have
opposite signs, the total will remain unchanged, as we are only swapping the positions
of the +1 and 1.
<o:p> </o:p>
(ii)
If we change the sign of cell
(X_{i} , Y_{i} ) where R_{i} and C_{i} have the
same sign, the total will either increase by 4 (if we go from two 1s to two
+1s) or decrease by 4.
<o:p> </o:p>
Thus, we can only ever change the total in
even amounts of 4.
<o:p> </o:p>
In the case of a 1997 x 1997 grid.
Σ_{i=1}(R_{i} + C_{i}) = 1997 + 1997
=
1997 * 2
= A number which
clearly does not have 4 as a factor, so we cannot lower the total to 0.

Posted by farcear
on 20100429 12:18:11 