The number of 5 digit numbers which are divisible by 3, using digits 1, 2, 3, 4, 5, 6, 7, 8 and 9 when repetition of digit is allowed is :

Options

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

Answer

To solve the problem of finding the number of 5-digit numbers divisible by 3 using the digits 1 through 9 (with repetition allowed), we can follow these steps:

### Step 1: Understanding the divisibility rule for 3
A number is divisible by 3 if the sum of its digits is divisible by 3. Therefore, we need to ensure that the sum of the digits in our 5-digit number meets this criterion.

**Hint:** Remember that the sum of the digits must be a multiple of 3.

### Step 2: Counting the choices for the first four digits
Since we can use any of the digits from 1 to 9 and repetition is allowed, each of the first four digits can be chosen in 9 different ways.

- Choices for the first digit: 9
- Choices for the second digit: 9
- Choices for the third digit: 9
- Choices for the fourth digit: 9

Thus, the total number of ways to choose the first four digits is:
\[ 9 \times 9 \times 9 \times 9 = 9^4 \]

**Hint:** Calculate \(9^4\) to find the total combinations for the first four digits.

### Step 3: Determining the last digit
The last digit must be chosen such that the total sum of all five digits is divisible by 3.

Let \(S\) be the sum of the first four digits. Depending on the value of \(S \mod 3\), we have three cases:

1. **Case 1:** If \(S \equiv 0 \mod 3\), then the last digit can be any digit that is also \(0 \mod 3\) (i.e., 3, 6, or 9). This gives us 3 options.
2. **Case 2:** If \(S \equiv 1 \mod 3\), then the last digit must be \(2 \mod 3\) (i.e., 2, 5, or 8). This also gives us 3 options.
3. **Case 3:** If \(S \equiv 2 \mod 3\), then the last digit must be \(1 \mod 3\) (i.e., 1, 4, or 7). This again gives us 3 options.

In all cases, there are 3 valid choices for the last digit.

**Hint:** Consider how the sum of the first four digits affects the choice of the last digit based on the modulus.

### Step 4: Combining the choices
Now, we can combine the choices for the first four digits and the last digit. The total number of valid 5-digit numbers is:
\[ 9^4 \times 3 \]

### Step 5: Calculating the final answer
Calculating \(9^4\):
\[ 9^4 = 6561 \]

Now, multiplying by the 3 choices for the last digit:
\[ 6561 \times 3 = 19683 \]

Thus, the total number of 5-digit numbers that can be formed using the digits 1 to 9, which are divisible by 3, is **19683**.

**Final Answer:** 19683