How many numbers can be made with digits 0, 5, 9 which are greater than 0 and less than a million?

To solve this problem, we need to consider the different cases for the number of digits in the number we are forming.

1. One-digit numbers: There are 2 possible numbers (5 and 9). Zero is not considered because the number must be greater than zero.

2. Two-digit numbers: The first digit can be either 5 or 9 (2 possibilities), and the second digit can be 0, 5, or 9 (3 possibilities). So there are 2*3 = 6 possible two-digit numbers.

3. Three-digit numbers: The first digit can be either 5 or 9 (2 possibilities), and the second and third digits can be 0, 5, or 9 (3 possibilities each). So there are 233 = 18 possible three-digit numbers.

4. Four-digit numbers: The first digit can be either 5 or 9 (2 possibilities), and the second, third, and fourth digits can be 0, 5, or 9 (3 possibilities each). So there are 233*3 = 54 possible four-digit numbers.

5. Five-digit numbers: The first digit can be either 5 or 9 (2 possibilities), and the second, third, fourth, and fifth digits can be 0, 5, or 9 (3 possibilities each). So there are 23333 = 162 possible five-digit numbers.

6. Six-digit numbers: The first digit can be either 5 or 9 (2 possibilities), and the second, third, fourth, fifth, and sixth digits can be 0, 5, or 9 (3 possibilities each). So there are 23333*3 = 486 possible six-digit numbers.

Adding up all these possibilities, we find that there are 2 + 6 + 18 + 54 + 162 + 486 = 728 numbers that can be made with the digits 0, 5, and 9 that are greater than 0 and less than a million.