Let n be an even positive integer. Write the numbers
1;2; ... ,n^2 in the squares of an nxn grid so that the
1st row, from left to right, is
1, 2, 3, ,...n and each row represents the continuation of the previous.
Color the numbers so that half of them
in each row and in each column are red and the other
half are black.
Prove that for each coloring, the sum of the red numbers equals to the sum of the black numbers.
Let's do something easier first.
Assume that each row is numbered 1,2,3,.... and paint the board red and black, so that half the numbers in each column are red and half black, ignoring rows. The red and black numbers on the whole board will obviously have the same sum (4 1's from the first column, 4 2's from the second column and so on)
What difference does the stipulated numbering add?
In the first row we have 1,2,...n; in the second (n+1), (n+2),...2n. etc. In the second row, we add n to each element, in the third row, 2n, etc. So we are adding exactly the same to every element in each row, resulting (almost too obviously for words) again in equality between red and black as far as the rows are concerned.
So, every colouring in accordance with the problem will produce equal totals of red and black numbers.
Edited on February 6, 2016, 2:40 am

Posted by broll
on 20160205 09:10:36 