In a certain chess position, each row and each column contains an odd number of pieces.

Prove that the total number of pieces on black squares is an even number.

Source: Proizvolov, “Problems Teaching Us How to Think".

Assuming the pieces have been legally placed, it must be possible to remove all but one from the first row, all but one from the second row..., such that each row and column contains just one piece.

This is equivalent to a board with pieces along a main diagonal for one colour, and a minor diagonal for the other (if the black pieces go on a1,b2,c3,.. then the white ones must start from either g8 or h7). As the diagonals approach, the only available square for the last piece ensures that parity is conserved, so there must be even numbers of both white and black pieces.

Edited on March 19, 2016, 1:36 am

Posted by broll on 2016-03-19 01:23:15