The squares of a rectangular chessboard with 10 rows and 14 columns are colored alternatingly black and white in the usual manner. Some stones are placed the board (possibly more than one on the same square) so that there are an odd number of stones in each row and each column. Show that the total number of stones on black squares is even.
Problem
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Tags: combinatorics proposed, combinatorics
Mscj
28.08.2018 13:05
There is an odd number of stones on each column and there are 14 columns Odd•14=even Therefore total stones=even Let a be the set of columns where the number of stones on black squares is odd Let c be the set of rows where the number of stones on black squares is odd Let d be the set of columns where the number of stones on black squares is even Let the intersection of a and c have the number of stones on black squares be even Then the intersection of a and d has an odd number of stones on black squares Whenever there is an odd number of stones on black squares in a row or column , there is an even number of stones on white squares Then the intersection of a and c have the number of stones on white squares be odd Then the intersection of a and d has an even number of stones on white squares Now I will find the parity of the number of stones in a N stones in a= Odd + even + odd +even = even Each column has an odd number of stones, so since a is even Number of columns in a is even Therefore n black stones in a is even, as it is the set of all columns with an odd number of stones on black N stones on black squares in b is even Therefore, total black stones = even +even =even Repeat for when the intersection of a and c have the number of stones on black squares being odd QED