Suppose p is even and q is odd. Then which of the following CANNOT be an integer?

Answer

To solve the problem, we need to analyze the expressions given the conditions that \( p \) is even and \( q \) is odd.

### Step-by-Step Solution:

1. **Understanding the Properties of Even and Odd Numbers**:
- An even number can be represented as \( p = 2n \) where \( n \) is an integer.
- An odd number can be represented as \( q = 2k + 1 \) where \( k \) is an integer.

2. **Analyzing the First Expression: \( \frac{p + q}{p} \)**:
- Substitute \( p \) and \( q \):
\[ p + q = 2n + (2k + 1) = 2n + 2k + 1 = 2(n + k) + 1 \]
- This shows that \( p + q \) is odd.
- Now, we divide by \( p \):
\[ \frac{p + q}{p} = \frac{2(n + k) + 1}{2n} \]
- Since \( p \) is even and \( p + q \) is odd, the result \( \frac{p + q}{p} \) will not yield an integer (an odd number divided by an even number cannot be an integer).

3. **Analyzing the Second Expression: \( \frac{pq}{3} \)**:
- Here, \( pq = (2n)(2k + 1) = 2n(2k + 1) = 4nk + 2n \).
- This product is even.
- Whether \( \frac{pq}{3} \) is an integer depends on whether \( pq \) is divisible by 3.
- If \( q \) (which is odd) happens to be a multiple of 3, then \( pq \) can be divisible by 3, making \( \frac{pq}{3} \) an integer. Thus, this expression **can** be an integer.

4. **Analyzing the Third Expression: \( \frac{q}{p^2} \)**:
- Here, \( p^2 \) is even (since the square of an even number is even).
- Thus, we have:
\[ \frac{q}{p^2} = \frac{2k + 1}{(2n)^2} = \frac{2k + 1}{4n^2} \]
- Since \( q \) is odd and \( p^2 \) is even, the result \( \frac{q}{p^2} \) will also not yield an integer (an odd number divided by an even number cannot be an integer).

### Conclusion:
The expressions that cannot be integers are:
- \( \frac{p + q}{p} \)
- \( \frac{q}{p^2} \)

Thus, the answer is that both \( \frac{p + q}{p} \) and \( \frac{q}{p^2} \) cannot be integers.

### Final Answer:
The expressions that cannot be integers are:
- \( \frac{p + q}{p} \)
- \( \frac{q}{p^2} \)