A five digit number divisible by 3 is to be formed using the digits 0,1,3,5,7,9 without repetitions. The total number of ways this can be done is

Options

A. 216
B. 192
C. 240
D. 3125

Answer

To solve the problem of forming a five-digit number divisible by 3 using the digits 0, 1, 3, 5, 7, and 9 without repetition, we can follow these steps:

### Step 1: Understand the divisibility rule for 3
A number is divisible by 3 if the sum of its digits is divisible by 3.

### Step 2: Calculate the total sum of available digits
The available digits are 0, 1, 3, 5, 7, and 9. We calculate their sum:
\[ 0 + 1 + 3 + 5 + 7 + 9 = 25 \]

### Step 3: Determine which digit to drop
Since we need to form a five-digit number, we must drop one digit. The sum of the remaining digits must be divisible by 3.

- If we drop **1**:
\[ 0 + 3 + 5 + 7 + 9 = 24 \quad (\text{divisible by } 3) \]

- If we drop **7**:
\[ 0 + 1 + 3 + 5 + 9 = 18 \quad (\text{divisible by } 3) \]

Thus, we can drop either 1 or 7.

### Step 4: Case 1 - Dropping 1
The remaining digits are 0, 3, 5, 7, and 9. We need to form a five-digit number.

- **First digit cannot be 0**: So, we have 4 choices (3, 5, 7, 9).
- **Second digit**: We can use any of the remaining 4 digits (including 0).
- **Third digit**: 3 choices left.
- **Fourth digit**: 2 choices left.
- **Fifth digit**: 1 choice left.

The total number of arrangements for this case is:
\[ 4 \times 4 \times 3 \times 2 \times 1 = 96 \]

### Step 5: Case 2 - Dropping 7
The remaining digits are 0, 1, 3, 5, and 9. We again need to form a five-digit number.

- **First digit cannot be 0**: So, we have 4 choices (1, 3, 5, 9).
- **Second digit**: We can use any of the remaining 4 digits (including 0).
- **Third digit**: 3 choices left.
- **Fourth digit**: 2 choices left.
- **Fifth digit**: 1 choice left.

The total number of arrangements for this case is also:
\[ 4 \times 4 \times 3 \times 2 \times 1 = 96 \]

### Step 6: Total arrangements
Adding the two cases together gives us:
\[ 96 + 96 = 192 \]

Thus, the total number of ways to form a five-digit number divisible by 3 using the digits 0, 1, 3, 5, 7, and 9 without repetition is **192**.