A five-digit number divisible by 3 is to be formed using the digits 0, 1, 2, 3, 4, and 5, without repetition. The total number of ways this can done is

Options

A. 216
B. 600
C. 240
D. 3125

Explanation

To solve the problem of forming a five-digit number divisible by 3 using the digits 0, 1, 2, 3, 4, and 5 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. We need to check the possible combinations of five digits from the given set.

### Step 2: Calculate the sum of all digits
The digits provided 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 using any combination of these digits, provided the sum of the selected digits is also divisible by 3.

### Step 3: Case Analysis
We will analyze two cases based on the digits we can use:

**Case 1: Exclude the digit 3**
- We use the digits 0, 1, 2, 4, and 5.
- The sum of these digits is:
\[ 0 + 1 + 2 + 4 + 5 = 12 \]
Since 12 is divisible by 3, any five-digit number formed with these digits will be divisible by 3.

**Step 4: Count the arrangements for Case 1**
- The first digit cannot be 0 (to ensure it remains a five-digit number). Thus, we can choose from 1, 2, 4, or 5 for the first position (4 options).
- After choosing the first digit, we have 4 remaining digits (including 0) for the second position (4 options).
- For the third position, we have 3 remaining digits (3 options).
- For the fourth position, we have 2 remaining digits (2 options).
- For the fifth position, we have 1 remaining digit (1 option).

The total arrangements for Case 1 can be calculated as:
\[ 4 \times 4 \times 3 \times 2 \times 1 = 96 \]

**Case 2: Include the digit 3**
- We use the digits 1, 2, 3, 4, and 5.
- The sum of these digits is:
\[ 1 + 2 + 3 + 4 + 5 = 15 \]
Since 15 is divisible by 3, any five-digit number formed with these digits will also be divisible by 3.

**Step 5: Count the arrangements for Case 2**
- In this case, we can use any of the digits for the first position (5 options).
- After choosing the first digit, we have 4 remaining digits for the second position (4 options).
- For the third position, we have 3 remaining digits (3 options).
- For the fourth position, we have 2 remaining digits (2 options).
- For the fifth position, we have 1 remaining digit (1 option).

The total arrangements for Case 2 can be calculated as:
\[ 5 \times 4 \times 3 \times 2 \times 1 = 120 \]

### Step 6: Total arrangements
Now, we sum the total arrangements from both cases:
\[ 96 + 120 = 216 \]

Thus, the total number of five-digit numbers that can be formed using the digits 0, 1, 2, 3, 4, and 5, without repetition, that are divisible by 3 is **216**.