We can understand this topic as putting the four numbers 0, 3, 6 and 9 in the above four brackets.

Look at the special requirements first: the requirements are odd, so only 3 or 9 can be put in the last bracket, so there are two ways.

The first bracket cannot be 0 (because it is the highest bit), so there are two numbers left (because the last bit has already occupied one), which means two possibilities.

There is no special requirement for the second and third brackets. In the second bracket, there are two possibilities besides the remaining two numbers in the 1 and the fourth bracket.

There is no choice for the third bracket, because there is only one number left, so put it in.

So, to sum up, it is: 2×2× 1×2=8 methods. The results are: 3069, 3609, 6039, 6309, 9063, 9603, 6903, 6093.

This is the classic "pigeon hole principle" in mathematics. I am an Olympic math teacher. Ask me questions!