Consider that the total available toy cars is 36.
The factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18, 36.
So Simon can make 1 pile of 36 toy cars, 2 piles of 18 cars each, 3 piles of 12 cars each, 4 piles of 9 cars each, 6 piles of 6 cars each, 9 piles of 4 cars each, 12 piles of 3 cars each, 18 piles of 2 cars each, and 36 piles of 1 car each.
Thus, the possible number of cars that could be in each pile are 1,2, 3, 4, 6, 9, 12, 18, 36.
Therefore, option D is the correct choice.
