If `10^(n) + 3.4^(n+2)+k` is divisible by `9, AA n in N` then the least positive integral value of k is

Options

A. (a) 1
B. (b) 3
C. (c) 5
D. (d) 7

Answer

To solve the problem, we need to find the least positive integral value of \( k \) such that the expression \( 10^n + 3 \cdot 4^{n+2} + k \) is divisible by 9 for all \( n \in \mathbb{N} \).

### Step-by-Step Solution:

1. **Define the Expression**:
Let \( P(n) = 10^n + 3 \cdot 4^{n+2} + k \).

2. **Substitute the Smallest Natural Number**:
We will start by substituting \( n = 1 \):
\[ P(1) = 10^1 + 3 \cdot 4^{1+2} + k = 10 + 3 \cdot 4^3 + k \]

3. **Calculate \( 4^3 \)**:
Calculate \( 4^3 \):
\[ 4^3 = 64 \]
Therefore,
\[ P(1) = 10 + 3 \cdot 64 + k = 10 + 192 + k = 202 + k \]

4. **Check for Divisibility by 9**:
We need \( 202 + k \) to be divisible by 9. To check this, we can find the sum of the digits of \( 202 + k \) and ensure it is a multiple of 9.

5. **Sum of the Digits**:
The sum of the digits of \( 202 \) is:
\[ 2 + 0 + 2 = 4 \]
Thus, the sum of the digits of \( 202 + k \) becomes:
\[ 4 + k \]

6. **Set Up the Equation**:
We want \( 4 + k \) to be equal to 9 (the smallest multiple of 9):
\[ 4 + k = 9 \]

7. **Solve for \( k \)**:
Rearranging gives:
\[ k = 9 - 4 = 5 \]

8. **Conclusion**:
The least positive integral value of \( k \) is \( 5 \).

### Final Answer:
The least positive integral value of \( k \) is \( \boxed{5} \).