A 4 by 4 board has two opposite corner squares removed, leaving 14 squares. Each domino covers exactly two squares that share an edge. Can 7 dominoes cover the board?
Text description of the figure
A 4 by 4 chessboard-style grid coloured in alternating light and dark squares, with the top-left and bottom-right squares removed. Both removed squares were the same colour.
Hints
Take them one at a time. The first gives nothing away.
1A nudge
Colour the board like a chessboard.
2The strategy
A domino always covers one light and one dark square, whatever way it is placed. Count how many of each colour the board has after the corners are removed.
3The full solution
The full board has 8 light and 8 dark squares. Opposite corners are the same colour, so removing them leaves 8 of one colour and 6 of the other. Seven dominoes would need 7 of each.
Solution
The reliable way
Colour the board like a chessboard. Whichever way a domino is laid, it covers exactly one light and one dark square, so seven dominoes must cover 7 light and 7 dark squares. The full board has 8 of each, and the two removed corners are the same colour because opposite corners always are. That leaves 8 of one colour and 6 of the other, which no set of dominoes can cover. So it is impossible. The transferable idea: choose a colouring that forces every piece to cover a fixed mixture, then count.
The elegant way
Every domino is colour-balanced, so any coverable region must be too; removing two same-coloured squares destroys that balance immediately.
Why this is on the test: Colouring arguments prove impossibility in one line where trial and error can run forever, and this is the canonical example.