, 2014

The key to the solution is the following observation.

(A) Each $1 \times 3$ rectangle contains exactly one black square.

Consider any $1 \times 3$ rectangle and let $r$ be the number of black squares it contains, then $0 \le r \le 3$ and we want to show $r = 1$.

Observe first that the two $3 \times 3$ squares adjacent to a $1 \times 3$ rectangle that contains $r$ black squares have to contain exactly $4 - r$ black squares. And also, the four $1 \times 3$ rectangles adjacent to a $3 \times 3$ square that contains exactly $4 - r$ black squares need to contain exactly $r$ black squares.

Let $0 \le a \le 9$ be the number of black squares among those that are marked with an asterisk. Then, there are $18r + 9 \cdot (4 - r) + a$ black squares in this $12 \times 12$ square. On the other hand, this $12 \times 12$ grid can be covered by twelve $3 \times 4$ rectangles and so it has to contain $12 \cdot 4 = 48$ black squares. Therefore, $18r + 9 \cdot (4 - r) + a = 48$, i.e. $9r + a = 12$. There is only one solution to this equation satisfying the constraints on $a$ and $r$, namely $r = 1$ and $a = 3$.

This shows that each $1 \times 3$ rectangle must contain exactly one black square.

(B) After choosing one square to be coloured black, there are exactly two possibilities to complete the colouring.

Consider the $3 \times 4$ rectangle with upper left corner the chosen black square. It follows from (A) that the squares $A$ and $B$ have to be black and that the remaining two black squares can only be among $C$, $D$, $E$, $F$.

Because of (A), either $C$ and $F$ or $D$ and $E$ are the black squares. Both patterns can be completed in a unique way, using (A) again, and it is easy to see that both satisfy the requirements of the problem:

(C) If we fix a $1 \times 3$ rectangle, there are three choices of the black square in it. For each of these choices we have seen in (B) that there are two ways to complete the colouring. Thus there are $6$ ways of colouring the grid.