Understanding the Problem
To find how many numbers less than 1000 are divisible by 5 formed from the digits 0 through 9, where each digit can be used only once, we need to consider the characteristics of such numbers.
Divisibility by 5
A number is divisible by 5 if its last digit is either 0 or 5. We will address both cases separately.
Case 1: Last Digit is 0
- The number can be a 1-digit, 2-digit, or 3-digit number.
- 1-digit number: Only the number 0 (which is not considered).
- 2-digit number: The first digit can be from {1, 2, 3, 4, 5, 6, 7, 8, 9}. Thus, there are 9 possibilities.
- 3-digit number: The first digit can be from {1, 2, 3, 4, 5, 6, 7, 8, 9} (9 choices). The second digit can be from the remaining 8 digits (0-9 excluding the first digit). Thus, we have:
- Choices for the first digit: 9
- Choices for the second digit: 8
- Total = 9 * 8 = 72
Case 2: Last Digit is 5
- Again, we will consider 1-digit, 2-digit, and 3-digit numbers.
- 1-digit number: Only the number 5 (1 possibility).
- 2-digit number: The first digit can be from {1, 2, 3, 4, 6, 7, 8, 9} (8 choices). Thus, there are 8 possibilities.
- 3-digit number: The first digit can be from {1, 2, 3, 4, 6, 7, 8, 9} (8 choices), and the second digit can be from the remaining 8 digits (excluding first and 5). Thus, we have:
- Choices for the first digit: 8
- Choices for the second digit: 8
- Total = 8 * 8 = 64
Final Calculation
- Total for last digit 0: 9 (2-digit) + 72 (3-digit) = 81
- Total for last digit 5: 1 (1-digit) + 8 (2-digit) + 64 (3-digit) = 73
Overall Total
- Total numbers = 81 + 73 = 154
Thus, the total number of numbers less than 1000 that can be formed with the digits 0-9, divisible by 5, and using each digit at most once is 154.