How many 5 digit numbers divisible by 3 can be formed using the digits 0, 1, 2, 3, 4 and 5, without repetition?

Options

A. 108
B. 216
C. 810
D. 180

Answer

To solve the problem of how many 5-digit numbers divisible by 3 can be formed using the digits 0, 1, 2, 3, 4, and 5 without repetition, we will 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 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, any combination of 5 digits from this set will also need to be checked for divisibility.

### Step 3: Determine the combinations of digits
We can form 5-digit numbers by excluding one digit at a time from the set of 6 digits. We will check which excluded digit keeps the sum of the remaining digits divisible by 3.

1. **Excluding 0**: Remaining digits are 1, 2, 3, 4, 5.
- Sum = 1 + 2 + 3 + 4 + 5 = 15 (divisible by 3)
- The first digit cannot be 0, so valid combinations can be formed.

2. **Excluding 1**: Remaining digits are 0, 2, 3, 4, 5.
- Sum = 0 + 2 + 3 + 4 + 5 = 14 (not divisible by 3)

3. **Excluding 2**: Remaining digits are 0, 1, 3, 4, 5.
- Sum = 0 + 1 + 3 + 4 + 5 = 13 (not divisible by 3)

4. **Excluding 3**: Remaining digits are 0, 1, 2, 4, 5.
- Sum = 0 + 1 + 2 + 4 + 5 = 12 (divisible by 3)
- The first digit cannot be 0, so valid combinations can be formed.

5. **Excluding 4**: Remaining digits are 0, 1, 2, 3, 5.
- Sum = 0 + 1 + 2 + 3 + 5 = 11 (not divisible by 3)

6. **Excluding 5**: Remaining digits are 0, 1, 2, 3, 4.
- Sum = 0 + 1 + 2 + 3 + 4 = 10 (not divisible by 3)

### Step 4: Count valid combinations
From the analysis, we have valid groups when excluding 0 and 3.

1. **Case 1: Excluding 0 (Digits: 1, 2, 3, 4, 5)**
- Total arrangements = 5! = 120

2. **Case 2: Excluding 3 (Digits: 0, 1, 2, 4, 5)**
- The first digit can be 1, 2, 4, or 5 (4 choices).
- After choosing the first digit, we have 4 remaining digits to arrange.
- Total arrangements = 4 (choices for the first digit) × 4! (arrangements of remaining digits) = 4 × 24 = 96

### Step 5: Total valid 5-digit numbers
Now, we add the valid combinations from both cases:
\[ 120 + 96 = 216 \]

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