IMSO essay 2018

Consider an $n \times n$ board. Label the rows $1,2,...,n$ and the columns $1,2,...,n$ . Place pawns freely in the $(n-1)\times(n-1)$ subsquare defined by rows $1,...,n-1$ and columns $1,...,n-1$ .

The colour of the pawn to be placed in the $n$ th column of each of the first $n-1$ rows is now fixed (we know the number of white pawns in the first $n-1$ columns of each row, and so the colour of the final pawn required to make the total number even is fixed --- if the number in the first $n-1$ columns is even, put a black pawn in the final column; otherwise put a white pawn in the final column).

Similarly, now that we know the colours of the pawns in the first $n-1$ rows, we can determine the colours of the pawns in the $n$ th row (to ensure that the number of white pawns in each of the $n$ columns is even).

There are now an even number of white pawns in each column, and so an even number of pawns in the whole grid. There are an even number of pawns in each of the first $n-1$ rows, and hence there must be an even number of pawns in the last row as well. The final condition is satisfied automatically.

Thus we have a uniquely defined placement of pawns (with an even number of white pawns in each row and column) for any placement of pawns in the $(n-1)\times(n-1)$ subsquare. Thus there are $2^{(n-1)^2}$ possible arrangements of pawns with the desired property. In the case $n=4$ , the answer is $2^9 = \boxed{512}$ , which is somewhat bigger than $140$ .