If `k^2 - 25` is an odd integer then which one of the following values gives the even number?

Options

A. a. `k^(2k - 1)`
B. b. `(3k - 3)(2k - 2)`
C. c. `(k + 5)(k -5 )`
D. d. any two of these

Answer

To solve the problem, we need to determine which value of \( k \) makes \( k^2 - 25 \) an even integer, given that \( k^2 - 25 \) is an odd integer.

### Step-by-Step Solution:

1. **Understanding the Condition**:
We know that \( k^2 - 25 \) is an odd integer. For the expression \( k^2 - 25 \) to be odd, \( k^2 \) must be even. This is because an even number minus an odd number (25 is odd) results in an odd number.

2. **Determining \( k \)**:
If \( k^2 \) is even, then \( k \) itself must be even (since the square of an odd number is odd). Therefore, \( k \) can take on even integer values.

3. **Evaluating the Options**:
We need to evaluate the options provided to find which one gives an even number when substituted into \( k^2 - 25 \).

Let's denote the options as:
- Option 1: \( k = 0 \)
- Option 2: \( k = 2 \)
- Option 3: \( k = 3 \)
- Option 4: \( k = 4 \)
- Option 5: \( k = 5 \)

4. **Calculating \( k^2 - 25 \) for Each Option**:
- For \( k = 0 \):
\[ k^2 - 25 = 0^2 - 25 = -25 \quad (\text{odd}) \]
- For \( k = 2 \):
\[ k^2 - 25 = 2^2 - 25 = 4 - 25 = -21 \quad (\text{odd}) \]
- For \( k = 3 \):
\[ k^2 - 25 = 3^2 - 25 = 9 - 25 = -16 \quad (\text{even}) \]
- For \( k = 4 \):
\[ k^2 - 25 = 4^2 - 25 = 16 - 25 = -9 \quad (\text{odd}) \]
- For \( k = 5 \):
\[ k^2 - 25 = 5^2 - 25 = 25 - 25 = 0 \quad (\text{even}) \]

5. **Identifying Even Results**:
From the calculations, we see that:
- \( k = 3 \) gives an even result (-16).
- \( k = 5 \) gives an even result (0).

6. **Conclusion**:
The values of \( k \) that give an even number when substituted into \( k^2 - 25 \) are \( k = 3 \) and \( k = 5 \). Therefore, the correct answer is that both options 3 and 5 yield even results.

### Final Answer:
The values of \( k \) that give an even number are \( k = 3 \) and \( k = 5 \).