Five digit number divisible by 3 is formed using 0 1 2 3 4 6 and 7 without repetition Total number of such numbers are

Explanation

To solve the problem of finding the total number of five-digit numbers divisible by 3 that can be formed using the digits 0, 1, 2, 3, 4, 6, and 7 without repetition, we will follow these steps:

### Step 1: Identify the digits and their sum
The available digits are: 0, 1, 2, 3, 4, 6, and 7.
To check for divisibility by 3, we need to find the sum of the digits.

Sum of all digits:
\[ 0 + 1 + 2 + 3 + 4 + 6 + 7 = 23 \]

### Step 2: Determine the groups of digits
We need to select 5 digits from the available 7 digits such that the sum of the selected digits is divisible by 3.

We can create groups of digits based on their sums:

1. **Group 1**: 0, 1, 2, 3, 6
- Sum = 0 + 1 + 2 + 3 + 6 = 12 (divisible by 3)

2. **Group 2**: 0, 1, 3, 4, 7
- Sum = 0 + 1 + 3 + 4 + 7 = 15 (divisible by 3)

3. **Group 3**: 0, 1, 4, 6, 7
- Sum = 0 + 1 + 4 + 6 + 7 = 18 (divisible by 3)

4. **Group 4**: 0, 2, 3, 4, 6
- Sum = 0 + 2 + 3 + 4 + 6 = 15 (divisible by 3)

5. **Group 5**: 1, 2, 3, 4, 6
- Sum = 1 + 2 + 3 + 4 + 6 = 16 (not divisible by 3)

The valid groups for our selection are Group 1, Group 2, Group 3, and Group 4.

### Step 3: Calculate the number of valid five-digit numbers from each group
1. **For Groups 1, 2, 3, and 4 (where 0 is included)**:
- The first digit cannot be 0. Therefore, we have to choose one of the other digits as the first digit.
- For each of these groups, we can choose the first digit in 4 ways (any digit except 0), then we have 4 remaining digits to arrange.

The number of arrangements for each group:
\[ \text{First digit choices} = 4 \quad (\text{cannot be 0}) \]
\[ \text{Remaining digits} = 4! = 24 \]
\[ \text{Total for each group} = 4 \times 24 = 96 \]
Since there are 4 such groups:
\[ \text{Total from Groups 1, 2, 3, and 4} = 4 \times 96 = 384 \]

2. **For Group 5 (where 0 is not included)**:
- All digits can be used, including 0, and any digit can be the first digit.
- The total arrangements for this group:
\[ 5! = 120 \]

### Step 4: Calculate the total number of five-digit numbers
Now, we add the totals from all groups:
\[ \text{Total} = 384 + 120 = 504 \]

### Final Answer
The total number of five-digit numbers that can be formed using the digits 0, 1, 2, 3, 4, 6, and 7 without repetition and that are divisible by 3 is **504**.

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