Guest Posted March 2, 2010 Report Share Posted March 2, 2010 (edited) On every square of a 1997x1997 checkerboard is written +1 or -1. Let, Ri = Product of all the numbers in the ith row, and: Ci = Product of all the numbers in the ith column. Prove that Sum (i = 1 to 1997) (Ri + Ci) is never equal to zero. Edited March 2, 2010 by K Sengupta Quote Link to comment Share on other sites More sharing options...
0 Guest Posted March 3, 2010 Report Share Posted March 3, 2010 Let S be the total sum, that is, S = Sum(i=1 to 1997)(Ri + Ci). What happens to S if we replace the number of just one square? Changing the number of square at row i and column j makes Ri and Cj change either from -1 to 1 or from 1 to -1. Hence Ri and Cj either decreases by 2, remains the same or increases by 2. Therefore, Ri + Cj changes by either -4, 0 or 4. Since all others rows and columns do not change, S changes by either -4, 0 or 4. In particular, S mod 4 is invariant. Now, if we change, one-by-one, all squares with -1, then we end up with a 1997x1997 chessboard filled up with 1s. For this configuration we have Ri = Ci = 1 for all i from 1 to 1997 and, thus, S = 2*1997 = 2 mod 4. Therefore, S for the initial configuration must be 2 mod 4. In particular, S cannot be 0. Quote Link to comment Share on other sites More sharing options...
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On every square of a 1997x1997 checkerboard is written +1 or -1.
Let, Ri = Product of all the numbers in the ith row, and:
Ci = Product of all the numbers in the ith column.
Prove that Sum (i = 1 to 1997) (Ri + Ci) is never equal to zero.
Edited by K SenguptaLink to comment
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