Show that a 444-by-77 chessboard can be completely tiled by 1-by-6 pieces but a 44-by-777 chessboard cannot.
I tried to show this by taking a multiplication table of both boards and trying to find the area that a tile covers.
This was not the right way to approach this problem. Anyone know how to do it?
HINT: Since $444$ is a multiple of $6$, it’s easy to tile the $444\times 77$ board. Color the $44\times 777$ board diagonally with $6$ colors, like this:
$$\begin{array}{|c|c|c|} \hline 1&2&3&4&5&6&1&2&3&\ldots\\ \hline 2&3&4&5&6&1&2&3&4&\ldots\\ \hline 3&4&5&6&1&2&3&4&5&\ldots\\ \hline 4&5&6&1&2&3&4&5&6&\ldots\\ \hline 5&6&1&2&3&4&5&6&1&\ldots\\ \hline 6&1&2&3&4&5&6&1&2&\ldots\\ \hline 1&2&3&4&5&6&1&2&3&\ldots\\ \hline 2&3&4&5&6&1&2&3&4&\ldots\\ \hline 3&4&5&6&1&2&3&4&5&\ldots\\ \hline \vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\ddots\\ \hline \end{array}$$
Each tile must cover exactly one square of each color. Show that this is impossible, because there are not the same numbers of squares of each color.