The number of 4 digit numbers made by using the digits 0, 1, 3, 5, 7, 9 (without repetition) and which are divisible by 3 are

Options

A. 144
B. 120
C. 216
D. none of these

Answer

To solve the problem of finding the number of 4-digit numbers made using the digits 0, 1, 3, 5, 7, and 9 (without repetition) that are divisible by 3, we can follow these steps:

### Step 1: Determine the sum of the digits
The digits available are 0, 1, 3, 5, 7, and 9. First, we need to calculate the sum of these digits:

\[ 0 + 1 + 3 + 5 + 7 + 9 = 25 \]

### Step 2: Check divisibility by 3
A number is divisible by 3 if the sum of its digits is divisible by 3. The total sum is 25, which is not divisible by 3. Therefore, we need to consider combinations of these digits whose sums are divisible by 3.

### Step 3: Possible sums
The possible sums of four digits from the given set should be less than or equal to 25 and divisible by 3. The closest sums we can consider are 24, 21, 18, 15, and 12.

### Step 4: Finding combinations
We will check which combinations of four digits can give us these sums:

1. **Sum = 24**:
- We can achieve this by excluding the digit 1 (0 + 3 + 5 + 7 + 9 = 24).
- Valid digits: 0, 3, 5, 7, 9

2. **Sum = 21**:
- We can achieve this by excluding the digit 3 (0 + 1 + 5 + 7 + 9 = 22).
- Valid digits: 0, 1, 5, 7, 9

3. **Sum = 18**:
- We can achieve this by excluding the digit 7 (0 + 1 + 3 + 5 + 9 = 18).
- Valid digits: 0, 1, 3, 5, 9

4. **Sum = 15**:
- We can achieve this by excluding the digit 9 (0 + 1 + 3 + 5 + 7 = 16).
- Valid digits: 0, 1, 3, 5, 7

5. **Sum = 12**:
- We can achieve this by excluding the digit 5 (0 + 1 + 3 + 7 + 9 = 20).
- Valid digits: 0, 1, 3, 7, 9

### Step 5: Arranging the digits
Now, we need to arrange the valid combinations of digits to form 4-digit numbers.

1. **For the sum of 24 (0, 3, 5, 7, 9)**:
- The first digit cannot be 0. So, we can choose from 3, 5, 7, or 9.
- If we choose 3 as the first digit, we can arrange the remaining three digits (0, 5, 7, 9) in \(3!\) ways.
- Total arrangements = \(3 \times 3! = 18\)

2. **For the sum of 21 (0, 1, 5, 7, 9)**:
- The first digit cannot be 0. So we can choose from 1, 5, 7, or 9.
- Total arrangements = \(4 \times 3! = 24\)

3. **For the sum of 18 (0, 1, 3, 5, 9)**:
- The first digit cannot be 0. So we can choose from 1, 3, 5, or 9.
- Total arrangements = \(4 \times 3! = 24\)

4. **For the sum of 15 (0, 1, 3, 5, 7)**:
- The first digit cannot be 0. So we can choose from 1, 3, 5, or 7.
- Total arrangements = \(4 \times 3! = 24\)

5. **For the sum of 12 (0, 1, 3, 7, 9)**:
- The first digit cannot be 0. So we can choose from 1, 3, 7, or 9.
- Total arrangements = \(4 \times 3! = 24\)

### Step 6: Total arrangements
Now, we sum up all the valid arrangements:

\[ 18 + 24 + 24 + 24 + 24 = 114 \]

### Final Answer
The total number of 4-digit numbers made by using the digits 0, 1, 3, 5, 7, and 9 (without repetition) that are divisible by 3 is **114**.

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