You have $2$ columns of $11$ squares in the middle, in the right and in the left you have columns of $9$ squares (centered on the ones of $11$ squares), then columns of $7,5,3,1$ squares. (This is the way it was explained in the original thread, http://www.artofproblemsolving.com/Forum/viewtopic.php?t=44430 ; anyway, i think you can understand how it looks) Several rooks stand on the table and beat all the squares ( a rook beats the square it stands in, too). Prove that one can remove several rooks such that not more than $11$ rooks are left and still beat all the table. Proposed by D. Rostovsky, based on folklore