Let p be a prime number strictly greater than 3. Then `p^2+17` will leave a remainder k, when divided by 12, the value of k is :

Options

A. 1
B. 5
C. 6
D. none of these

Answer

To solve the problem, we need to find the value of \( k \) when \( p^2 + 17 \) is divided by 12, where \( p \) is a prime number greater than 3.

### Step-by-Step Solution:

1. **Understanding the properties of prime numbers greater than 3**:
- Any prime number \( p > 3 \) can be expressed in the form of \( 6n \pm 1 \) for some integer \( n \). This is because all primes greater than 3 are either one more or one less than a multiple of 6.

2. **Calculating \( p^2 \)**:
- If \( p = 6n + 1 \), then:
\[ p^2 = (6n + 1)^2 = 36n^2 + 12n + 1 \]
- If \( p = 6n - 1 \), then:
\[ p^2 = (6n - 1)^2 = 36n^2 - 12n + 1 \]
- In both cases, \( p^2 \) can be simplified modulo 12:
\[ p^2 \equiv 1 \mod 12 \]

3. **Adding 17 to \( p^2 \)**:
- Now, we need to compute \( p^2 + 17 \):
\[ p^2 + 17 \equiv 1 + 17 \mod 12 \]
- Simplifying \( 1 + 17 \):
\[ 1 + 17 = 18 \]
- Now, we find \( 18 \mod 12 \):
\[ 18 \div 12 = 1 \quad \text{(remainder 6)} \]
- Therefore:
\[ 18 \equiv 6 \mod 12 \]

4. **Conclusion**:
- The remainder \( k \) when \( p^2 + 17 \) is divided by 12 is:
\[ k = 6 \]

### Final Answer:
The value of \( k \) is \( 6 \).