To find the number of five-digit numbers formed with the digits 0, 1, 2, 3, 4, and 5 (without repetition) that are divisible by 6, we need to ensure that the numbers meet the criteria for divisibility by both 2 and 3.
### Step-by-step Solution:
1. **Identify the Criteria for Divisibility by 6**:
- A number is divisible by 6 if it is divisible by both 2 and 3.
- For divisibility by 2: The last digit must be even (0, 2, or 4).
- For divisibility by 3: The sum of the digits must be divisible by 3.
2. **Calculate the Total Sum of Available Digits**:
- The digits available are 0, 1, 2, 3, 4, and 5.
- The sum of these digits is \(0 + 1 + 2 + 3 + 4 + 5 = 15\).
- Since 15 is divisible by 3, any combination of 5 digits from this set will also be divisible by 3.
3. **Form Groups of Digits**:
- We can form two groups of five digits:
- Group 1: {1, 2, 3, 4, 5}
- Group 2: {0, 1, 2, 4, 5}
- Both groups have a sum that is divisible by 3.
4. **Check for Last Digit Options**:
- For both groups, the last digit must be even (0, 2, or 4).
5. **Calculating for Group 1: {1, 2, 3, 4, 5}**:
- Possible last digits: 2 or 4.
- **Case 1: Last digit is 2**:
- Remaining digits: {1, 3, 4, 5}
- The first digit can be any of these 4 digits (1, 3, 4, or 5).
- The number of arrangements = \(4 \times 3 \times 2 \times 1 = 24\).
- **Case 2: Last digit is 4**:
- Remaining digits: {1, 2, 3, 5}
- The first digit can be any of these 4 digits (1, 2, 3, or 5).
- The number of arrangements = \(4 \times 3 \times 2 \times 1 = 24\).
- Total for Group 1 = \(24 + 24 = 48\).
6. **Calculating for Group 2: {0, 1, 2, 4, 5}**:
- Possible last digits: 0, 2, or 4.
- **Case 1: Last digit is 0**:
- Remaining digits: {1, 2, 4, 5}
- The first digit can be any of these 4 digits (1, 2, 4, or 5).
- The number of arrangements = \(4 \times 3 \times 2 \times 1 = 24\).
- **Case 2: Last digit is 2**:
- Remaining digits: {0, 1, 4, 5}
- The first digit can be any of 1, 4, or 5 (0 cannot be the first digit).
- The number of arrangements = \(3 \times 3 \times 2 \times 1 = 18\).
- **Case 3: Last digit is 4**:
- Remaining digits: {0, 1, 2, 5}
- The first digit can be any of 1, 2, or 5 (0 cannot be the first digit).
- The number of arrangements = \(3 \times 3 \times 2 \times 1 = 18\).
- Total for Group 2 = \(24 + 18 + 18 = 60\).
7. **Final Calculation**:
- Total number of valid five-digit numbers = Total from Group 1 + Total from Group 2 = \(48 + 60 = 108\).
### Conclusion:
The total number of five-digit numbers formed with the digits 0, 1, 2, 3, 4, and 5 (without repetition) that are divisible by 6 is **108**.
