If (`12^n` + 1) is divisible by 13, then n is

Options

A. any even integer only
B. any integer only
C. any odd integer
D. only 12

Answer

To solve the problem, we need to determine the values of \( n \) for which \( 12^n + 1 \) is divisible by 13.

### Step-by-Step Solution:

1. **Understanding the Expression**: We start with the expression \( 12^n + 1 \). We want to find out when this expression is divisible by 13.

2. **Using Modular Arithmetic**: We can analyze \( 12^n + 1 \) modulo 13. We need to find \( 12^n + 1 \equiv 0 \mod 13 \).

3. **Simplifying the Base**: Notice that \( 12 \equiv -1 \mod 13 \). Therefore, we can rewrite \( 12^n \) as:
\[ 12^n \equiv (-1)^n \mod 13 \]

4. **Substituting Back**: Now substituting back into our expression, we have:
\[ 12^n + 1 \equiv (-1)^n + 1 \mod 13 \]

5. **Analyzing Cases**: We need to analyze two cases based on whether \( n \) is even or odd:
- **Case 1**: If \( n \) is even, then \( (-1)^n = 1 \):
\[ (-1)^n + 1 \equiv 1 + 1 \equiv 2 \mod 13 \]
This is not divisible by 13.

- **Case 2**: If \( n \) is odd, then \( (-1)^n = -1 \):
\[ (-1)^n + 1 \equiv -1 + 1 \equiv 0 \mod 13 \]
This is divisible by 13.

6. **Conclusion**: From our analysis, we conclude that \( 12^n + 1 \) is divisible by 13 only when \( n \) is an odd integer.

### Final Answer:
Thus, the value of \( n \) must be any odd integer.