 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Fifty-fifty (Posted on 2016-02-05) Let n be an even positive integer. Write the numbers 1;2; ... ,n^2 in the squares of an nxn 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. Comments: ( Back to comment list | You must be logged in to post comments.) Solution Comment 2 of 2 | Change each number to an ordered pair by subtracting 1 and converting to base n.  Example: if n=8 then 1=(0,0), 2=(0,1), ..., 8=(0,7), 9=(1,0), 10=(1,1), ..., 64=(7,7).

We are given that in any row half the numbers are black and half are red.  The same condition holds for the columns.

In each row the first coordinate is the same, (0,x) in row 1, (1,x) in row 2, ... (n-1,x) in row n.  Due to the coloring exactly half of the 0's, 1's, ... (n-1)'s are red and half are black.  So the sum of the first coordinate of all the black squares equals the sum for the red squares.

The columns are handled similarly for the second coordinate, coming to the conclusion that the sum of the second coordinate of all the black squares equals the sum for the red squares.

Combining both the first and second coordinates means that the sum of all the red coordinates equals the sum of all the black coordinates, which then implies the sum of all the original red numbers equals the sum of all the original black numbers. QED.

 Posted by Brian Smith on 2016-02-05 09:53:41 Please log in:

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