On a checkered square $10 \times 10$ the cells of the upper left $5 \times 5$ square are black and all the other cells are white. What is the maximal $n$ such that the original square can be dissected (along the borders of the cells) into $n$ polygons such that in each of them the number of black cells is three times less than the number of white cells? (The polygons need not be congruent or even equal in area.)
Problem
Source: 2016 Saudi Arabia BMO TST , level 4, I p4
Tags: combinatorics, combinatorial geometry, table