The number of five digit numbers formed with the digits 0, 1, 2, 3, 4 and 5 (without repetition) and divisible by 6 are

Options

A. 72
B. 84
C. 96
D. 108

Answer

To solve the problem of finding 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 follow these steps:

### Step 1: Understand the divisibility rules for 6
A number is divisible by 6 if it is divisible by both 2 and 3.

- **Divisibility by 2**: The last digit must be even. The even digits available are 0, 2, and 4.
- **Divisibility by 3**: The sum of the digits must be divisible by 3.

### Step 2: Calculate the sum of the digits
The digits we have 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, we can form five-digit numbers by leaving out one digit, ensuring that the sum of the remaining digits is still divisible by 3.

### Step 3: Determine which digit to leave out
We will analyze cases based on which digit we leave out:

1. **Leave out 0**: Remaining digits are 1, 2, 3, 4, 5. The sum is \(1 + 2 + 3 + 4 + 5 = 15\) (divisible by 3).
2. **Leave out 3**: Remaining digits are 0, 1, 2, 4, 5. The sum is \(0 + 1 + 2 + 4 + 5 = 12\) (divisible by 3).
3. **Leave out 1**: Remaining digits are 0, 2, 3, 4, 5. The sum is \(0 + 2 + 3 + 4 + 5 = 14\) (not divisible by 3).
4. **Leave out 2**: Remaining digits are 0, 1, 3, 4, 5. The sum is \(0 + 1 + 3 + 4 + 5 = 13\) (not divisible by 3).
5. **Leave out 4**: Remaining digits are 0, 1, 2, 3, 5. The sum is \(0 + 1 + 2 + 3 + 5 = 11\) (not divisible by 3).
6. **Leave out 5**: Remaining digits are 0, 1, 2, 3, 4. The sum is \(0 + 1 + 2 + 3 + 4 = 10\) (not divisible by 3).

Thus, the valid cases are:
- Case 1: Leave out 0 (digits 1, 2, 3, 4, 5)
- Case 2: Leave out 3 (digits 0, 1, 2, 4, 5)

### Step 4: Count the valid five-digit numbers for each case

#### Case 1: Leaving out 0 (Digits: 1, 2, 3, 4, 5)
- The last digit can be 2 or 4 (to satisfy divisibility by 2).

1. **Last digit = 2**:
- Remaining digits: 1, 3, 4, 5
- Number of arrangements = \(4! = 24\)

2. **Last digit = 4**:
- Remaining digits: 1, 2, 3, 5
- Number of arrangements = \(4! = 24\)

Total for Case 1 = \(24 + 24 = 48\)

#### Case 2: Leaving out 3 (Digits: 0, 1, 2, 4, 5)
- The last digit can be 0, 2, or 4.

1. **Last digit = 0**:
- Remaining digits: 1, 2, 4, 5
- Number of arrangements = \(4! = 24\)

2. **Last digit = 2**:
- Remaining digits: 0, 1, 4, 5 (0 cannot be the first digit)
- Number of valid arrangements = \(3! \times 3 = 18\) (3 choices for the first digit)

3. **Last digit = 4**:
- Remaining digits: 0, 1, 2, 5 (0 cannot be the first digit)
- Number of valid arrangements = \(3! \times 3 = 18\) (3 choices for the first digit)

Total for Case 2 = \(24 + 18 + 18 = 60\)

### Step 5: Combine the results
Total valid five-digit numbers = Case 1 + Case 2 = \(48 + 60 = 108\)

Thus, the total number of five-digit numbers that can be formed with the digits 0, 1, 2, 3, 4, and 5 (without repetition) that are divisible by 6 is **108**.