Suppose that $p$ and $q$ are two different prime numbers and that $n=p^{2} q^{2}$. The number of possible values of $n$ with $n<1000$ is
(A) 5
(B) 6
(C) 4
(D) 8
(E) 7
We note that $n=p^{2} q^{2}=(p q)^{2}$.

Since $n<1000$, then $(p q)^{2}<1000$ and so $p q<\sqrt{1000} \approx 31.6$.

Finding the number of possible values of $n$ is thus equivalent to finding the number of positive integers $m$ with $1 \leq m \leq 31<\sqrt{1000}$ that are the product of two prime numbers.

The prime numbers that are at most 31 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31.

The distinct products of pairs of these that are at most 31 are:

$$
\begin{gathered}
2 \times 3=6 \quad 2 \times 5=10 \quad 2 \times 7=14 \quad 2 \times 11=22 \quad 2 \times 13=26 \\
3 \times 5=15 \quad 3 \times 7=21
\end{gathered}
$$

Any other product either duplicates one that we have counted already, or is larger than 31. Therefore, there are 7 such values of $n$.

ANswer: (E)