Let `k` be a positive integer such that `(k+ 4)` is divisible by `7`. Let `n` be the smallest positive integer `gt 2`, and if `(k+ n^(2))` is divisible by `7`, then `n` equals.

Options

A. 3
B. 5
C. 7
D. 9

Answer

To solve the problem step by step, we need to follow the conditions given in the question.

### Step 1: Understand the conditions
We know that \( k + 4 \) is divisible by \( 7 \). This means:
\[ k + 4 \equiv 0 \mod 7 \]
or equivalently,
\[ k \equiv -4 \mod 7 \]
Since \(-4\) can be expressed as \(3\) in modulo \(7\) (because \(-4 + 7 = 3\)), we can say:
\[ k \equiv 3 \mod 7 \]

### Step 2: Set up the second condition
Next, we need to find the smallest positive integer \( n > 2 \) such that \( k + n^2 \) is also divisible by \( 7 \):
\[ k + n^2 \equiv 0 \mod 7 \]
This can be rewritten using our earlier result:
\[ 3 + n^2 \equiv 0 \mod 7 \]
or
\[ n^2 \equiv -3 \mod 7 \]
which is equivalent to:
\[ n^2 \equiv 4 \mod 7 \]

### Step 3: Find values of \( n \)
Now we need to find the smallest positive integer \( n > 2 \) such that \( n^2 \equiv 4 \mod 7 \). We will check integers starting from \( n = 3 \):

- For \( n = 3 \):
\[ n^2 = 3^2 = 9 \equiv 2 \mod 7 \quad (\text{not } 4) \]

- For \( n = 4 \):
\[ n^2 = 4^2 = 16 \equiv 2 \mod 7 \quad (\text{not } 4) \]

- For \( n = 5 \):
\[ n^2 = 5^2 = 25 \equiv 4 \mod 7 \quad (\text{this works!}) \]

### Conclusion
The smallest positive integer \( n > 2 \) such that \( k + n^2 \) is divisible by \( 7 \) is:
\[ \boxed{5} \]