A $23\times23$ square is completely tiled by $1\times1, 2\times2$ and $3\times3$ tiles. What is the smallest number of $1\times1$ tiles needed?
This is the solution
If we color the rows of the $23\times23$ square alternating white and black (starting with white), there will be exactly $23$ more white squares than black squares. On the other hand, $2\times2$ squares cover $2$ of each color, while $3\times 3$ squares cover $6$ of one color and $3$ of the other. So the square cannot be filled without using at least one $1\times1$ square (otherwise, the difference between number of white squares and number of black squares would be a multiple of $3$).
I can't understand. $3\times3$ tile according to me will cover $5$ of one color and $4$ of other.
Please explain.