# Combinatorial Agriculture

You have just acquired a $64$ acre farm, in the shape of a square, and divided into an eight by eight array of one acre subplots. You have $21$ crops to plant. Each crop requires its own $3$ acre plot of land, which must consist of three subplots in the shape of a three by one rectangle. Once the crops have been planted, there will be $64-21\times 3=1$ subplot leftover, where you will build your farmhouse.

It isn't the season for planting yet, so you plan to build your farmhouse now, then plant the crops later. Which subplots can you build your farmhouse on which still allow you to plant all of your crops?

It is only possible in

$4$ cells

Those cells are

The black ones in the bottom grid.

Because

If you tile the grid with three colours, following the above pattern (the right one is symmetrical to the left one), you get 21 green, 21 blue and 22 red cells.
Though, any 3x1 rectangle occupies exactly 1 red, 1 blue and 1 green cell, so the farmhouse must be in a red cell.
Also, it must be in a red cell in both grids (otherwise it would violate the above condition in one grid), and if you do the intersection of red cells (the cells that are red in both grids) you only get the 4 black cells in the bottom grid.

An example of valid configuration is

• Can you explain why it has to be a red cell that lies in both of those grids. It makes sense that it has to be one of those red cells (from either grid) but could you give an explanation for the correct spaces needing to intersect both grids. – MisterEman22 May 17 '15 at 19:22
• @MisterEman22 Sure! Suppose that the farmhouse lies on a red cell of ONLY one grid. Then, in the second grid the same farmhouse would be in a green/blue cell, and that creates an impossible scenario (22 red cells left, 21 green and 20 blue). – leoll2 May 17 '15 at 19:26
• Okay I got it now. I was confusing myself with including the other 2 possible orientations of the grid. They would be with blue and green switched on both of the grids you have now, which wouldn't matter because all of the reds would be in the same spots. – MisterEman22 May 17 '15 at 20:11
• This answer is optimal in every way! – Mike Earnest May 17 '15 at 23:44

Here is one possibility:

This could be rotated to allow for the 4 plots that are in that position (3rd plot from the two closest edges). So far I haven't been able to find any more. Still trying to figure some mathematical reason for this.