The number of five-digit numbers which are divisible by `3` that can be formed by using the digits `1,2,3,4,5,6,7,8` and `9`, when repetition of digits is allowed, is

Options

A. `3^(9)`
B. `4.3^(8)`
C. `5.3^(8)`
D. `7.3^(8)`

Explanation

To solve the problem of finding the number of five-digit numbers divisible by 3 that can be formed using the digits 1 through 9 (with repetition allowed), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Divisibility Rule for 3**:
A number is divisible by 3 if the sum of its digits is divisible by 3.

2. **Identifying the Available Digits**:
The digits available to us are 1, 2, 3, 4, 5, 6, 7, 8, and 9. There are a total of 9 digits.

3. **Calculating the Total Combinations**:
Since repetition of digits is allowed, for each of the five positions in the five-digit number, we can choose any of the 9 digits. Therefore, the total number of ways to fill the first four positions is:
\[ 9 \times 9 \times 9 \times 9 = 9^4 \]

4. **Considering the Last Digit Based on the Sum**:
The last digit must be chosen based on the sum of the first four digits to ensure the total sum is divisible by 3.

- **Case 1**: If the sum of the first four digits is already divisible by 3 (i.e., \(3n\)), we can choose any digit that is also divisible by 3. The digits that are divisible by 3 from our set are 3, 6, and 9. Thus, we have 3 options for the last digit.

- **Case 2**: If the sum of the first four digits gives a remainder of 1 when divided by 3 (i.e., \(3n + 1\)), we need to choose a digit that, when added, makes the total divisible by 3. The digits that can be added are 2, 5, and 8. Thus, we again have 3 options for the last digit.

- **Case 3**: If the sum of the first four digits gives a remainder of 2 when divided by 3 (i.e., \(3n + 2\)), we need to choose a digit that makes the total divisible by 3. The digits that can be added are 1, 4, and 7. Thus, we again have 3 options for the last digit.

5. **Calculating the Total Valid Combinations**:
Since there are 3 options for the last digit in each case, regardless of the case, the total number of five-digit numbers divisible by 3 is:
\[ 9^4 \times 3 \]

6. **Final Calculation**:
Now we compute \(9^4\):
\[ 9^4 = 6561 \]
Therefore, the total number of five-digit numbers divisible by 3 is:
\[ 6561 \times 3 = 19683 \]

### Final Answer:
The total number of five-digit numbers that can be formed using the digits 1 to 9, which are divisible by 3, is **19683**.