MHB Problem of the Week # 277 - Aug 22, 2017

[Ackbach]

Here is this week's POTW:

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Let $n$ be an even positive integer. Write the numbers $1,2,\ldots,n^2$ in the squares of an $n\times n$ grid so that the $k$-th row, from left to right, is \[(k-1)n+1,(k-1)n+2,\ldots, (k-1)n+n.\]
Color the squares of the grid so that half of the squares in each row and in each column are red and the other half are black (a checkerboard coloring is one possibility). Prove that for each coloring, the sum of the numbers on the red squares is equal to the sum of the numbers on the black squares.

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[Ackbach]

Re: Problem Of The Week # 277 - Aug 22, 2017

This was Problem B-1 in the 2001 William Lowell Putnam Mathematical Competition.

Congratulations to Opalg for his correct solution, which follows:

[sp]For a given colouring, let $R$ be the sum of the numbers on the red squares and let $B$ be the sum of the numbers on the black squares.

Suppose that we change the numbers by subtracting $(k-1)n$ from each element in the $k$th row, for every $k$ ($1\leqslant k\leqslant n$). The values of $R$ and $B$ will change. But since each row contains equal numbers of red and black squares, $R$ and $B$ will have changed by equal amounts. So the difference $R-B$ will be the same as before.

In the altered grid, each row consists of the numbers $1,2,\ldots,n$. Thus for each $k$, the $k$th column will consist entirely of $k$s. Since each column contains equal numbers of red and black squares, the sum of the numbers on the red squares in column $k$ is equal to the sum of the numbers on the black squares of that column, namely $\frac12kn$. Therefore $R=B$ (in the altered grid), and so $R=B$ for the original grid.[/sp]