lean_workbook_plus_48042

Let's number the board from the top leftmost cell to the bottom rightmost cell with numbers from $0$ to $99$ . The number in cell $(x, y)$ will be equal to $10\cdot x + y$ . Each domino will now cover 3 squares that the sum of the numbers in them is divisible by $3$ . Since the sum of all the numbers is also divisible by $3$ , the number in the leftover cell must be divisible by $3$ . Now, each $90^\circ$ rotation with turn the number in cell $(x,y)$ from $10\cdot x + y$ to $10\cdot y + (9 - x)$ . If the cell $(x,y)$ is leftover in the first configuration, then it is still a leftover cell after the rotation. Thus, we have $3\mid 10\cdot x + y$ and $3\mid 10\cdot y + (9 - x)$ , or $3\mid x$ and $3\mid y$ . Since $0\le x, y\le 9$ , so $x,y\in\{0, 3, 6, 9\}$ . There are $16$ such pairs, account for $16$ cells.