TITLE: Is it possible to paint a $19\times19$ board so that every $10\times10$ square has a different number of colored squares?
QUESTION [5 upvotes]: Is it possible to paint a $19\times19$ board so that every $10\times10$ square has a different number of colored squares?
I'm not quite sure how to definitely prove this. What I've done so far is figure out that there's a $100$ different $10\times10$ squares in a $19\times19$ board. This means that we'll have one $10\times10$ square for every number from $1$ to $100$ or $0$ to $99$. I've thought about looking at the square with $0$ or $1$ colored squares and seeing what it'll mean for the other squares. I know that the $0$ and $99$ or $1$ and $100$ squares must share exactly one square. If they share anymore it'll be impossible. Not sure where to go from here, any help is appreciated.

REPLY [10 votes]: Motivated by the fact that the $10 \times 10$ squares with $0$ and $99$  (or $1$ and $100$) colored squares share exactly one square, I feel that this construction is quite natural:

(The blue square has $0$ colored squares, while the red square has $99$ colored squares.)
It is not hard to verify that this construction satisfies the requirement. In fact, let $S(i, j)$ denote the $10 \times 10$ square whose top left square is at the $i^{\text{th}}$ row and $j^{\text{th}}$ column. (Here, we start counting from zero.) Then, $S(i, j)$ has exactly $10i + j$ colored squares.