Problem
COM-B2-M05-P014 Odd Rows and Columns
On a \(10\times14\) board coloured like a chessboard, some cells are marked. In every row and every column, an odd number of cells is marked. Prove that the number of marked black cells is even.
Express the colour of a cell through the parity of the sum of its row and column indices.
Number the rows \(1,\ldots,10\) and the columns \(1,\ldots,14\). Let black cells be those for which \(i+j\) is odd. Let \(x_{ij}\) be \(0\) or \(1\), indicating whether the cell is marked.
Modulo \(2\), the number of marked black cells is
\[\sum_{i,j}(i+j)x_{ij}=\sum_i i\sum_j x_{ij}+\sum_j j\sum_i x_{ij}.\]
In every row and every column the number of marks is odd, so each such sum is \(1\) modulo \(2\). Therefore the desired parity is
\[\sum_{i=1}^{10} i+\sum_{j=1}^{14} j=55+105=160\equiv 0\pmod 2.\]
Thus the number of marked black cells is even.
The idea is inspired by an olympiad method using parity sums on a chessboard; the dimensions and wording are changed.