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

Options

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

Answer

To find the value of \( k \) when \( p^2 + 17 \) is divided by 12, where \( p \) is a prime number strictly greater than 3, we can follow these steps:

### Step 1: Understand the properties of prime numbers greater than 3
All prime numbers greater than 3 are either of the form \( 6n + 1 \) or \( 6n + 5 \) (which is equivalent to \( 6n - 1 \)). This means we can express any prime \( p > 3 \) in these two forms.

### Step 2: Calculate \( p^2 \) for both forms
1. **For \( p = 6n + 1 \)**:
\[ p^2 = (6n + 1)^2 = 36n^2 + 12n + 1 \]
When divided by 12, we can simplify:
\[ p^2 \mod 12 = (36n^2 + 12n + 1) \mod 12 = 1 \]

2. **For \( p = 6n + 5 \)**:
\[ p^2 = (6n + 5)^2 = 36n^2 + 60n + 25 \]
When divided by 12, we simplify:
\[ p^2 \mod 12 = (36n^2 + 60n + 25) \mod 12 = 25 \mod 12 = 1 \]

### Step 3: Add 17 to \( p^2 \)
Now, we add 17 to both cases:
1. **For \( p = 6n + 1 \)**:
\[ p^2 + 17 \equiv 1 + 17 \mod 12 = 18 \mod 12 = 6 \]

2. **For \( p = 6n + 5 \)**:
\[ p^2 + 17 \equiv 1 + 17 \mod 12 = 18 \mod 12 = 6 \]

### Conclusion
In both cases, we find that:
\[ p^2 + 17 \equiv 6 \mod 12 \]
Thus, the value of \( k \) is \( 6 \).

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

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