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 to 9 with repetition allowed, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Divisibility by 3**:
A number is divisible by 3 if the sum of its digits is divisible by 3. Hence, we need to ensure that the sum of the five digits we choose is divisible by 3.

2. **Identifying the Digits**:
The digits available are 1, 2, 3, 4, 5, 6, 7, 8, and 9. The sum of these digits modulo 3 can be classified as follows:
- Digits giving remainder 0 when divided by 3: 3, 6, 9
- Digits giving remainder 1 when divided by 3: 1, 4, 7
- Digits giving remainder 2 when divided by 3: 2, 5, 8

3. **Counting the Total Combinations**:
Since repetition of digits is allowed, each of the five positions in the number can be filled by any of the 9 digits. Therefore, the total number of five-digit combinations (without considering divisibility) is:
\[ 9^5 \]

4. **Classifying the Sums**:
We need to classify the sums of the digits based on their remainders when divided by 3. The possible sums can be:
- \(3n\) (divisible by 3)
- \(3n + 1\)
- \(3n + 2\)

5. **Calculating the Valid Combinations**:
- For a sum that is \(3n\): We can choose any of the digits (3, 6, 9) for the last digit to ensure the sum is divisible by 3. The number of ways to choose the last digit is 3, and for the first four digits, we can choose any of the 9 digits. Thus, the total combinations for this case is:
\[ 9^4 \times 3 \]
- For a sum that is \(3n + 1\): The last digit must be one of (1, 4, 7) to make the total sum divisible by 3. The number of ways to choose the last digit is 3, and for the first four digits, we can choose any of the 9 digits. Thus, the total combinations for this case is:
\[ 9^4 \times 3 \]
- For a sum that is \(3n + 2\): The last digit must be one of (2, 5, 8). The number of ways to choose the last digit is 3, and for the first four digits, we can choose any of the 9 digits. Thus, the total combinations for this case is:
\[ 9^4 \times 3 \]

6. **Combining the Cases**:
The total number of five-digit numbers divisible by 3 is the sum of the valid combinations from all three cases:
\[ \text{Total} = 9^4 \times 3 + 9^4 \times 3 + 9^4 \times 3 = 3 \times 9^4 \times 3 = 9^4 \times 9 = 9^5 \]

7. **Final Calculation**:
Since \(9^5 = 59049\), the total number of five-digit numbers that can be formed using the digits 1 to 9 and are divisible by 3 is:
\[ \boxed{59049} \]