We denote the total number of the white cells with 1 by X.Let C be the matrix which represents the 0,1's on the chessboard.Let H_a be the sum of the a-th row of C and L_b be the sum of the b-th column of C.Consider the sumWe claim S is the sum of all white 1's with +2 or -2 in front of them. This is because the white cells correspond to the coefficients of C whose column number and row number are both even or both odd. Black cells cancel out when we form the sum because they correspond to coefficients whose column number and row number are different modulo 2.Its easy to see that S = 2*X (mode 4) . And from the condition given by the problem, it is very easy to see S = 0 (mode 4) (To prove this, we only need to use are all odd as stated in the problem and the equality which is obviously true given the definitions of ). Hence X is even.
Last edited by pk14
on December 23rd, 2007, 11:00 pm, edited 1 time in total.