$30 \cdot 33$ five natural numbers, each of which has no prime factor greater than 3. Then there must be two natural numbers that are 
(A) the sum of perfect squares.
(B) the difference of perfect squares.
(C) the product of perfect squares.
(D) the quotient of perfect squares.
(2nd "Five Sheep Cup" Junior High School Mathematics Competition, 1990)
[Solution] According to the problem, these five natural numbers can be expressed as
$a_{k}=2^{m_{k}} k_{k} \cdot 3^{n_{k}}, \quad k=1,2,3,4,5$, where $m_{k}, n_{k}$ are non-negative integers.
The pairs $\left(m_{k}, n_{k}\right)$ have four possible combinations: (odd, odd), (odd, even), (even, odd), (even, even). Therefore, among the five natural numbers, there must be two with the same parity for their exponents, without loss of generality, let $m_{1}, m_{2}$ and $n_{1}, n_{2}$ be of the same parity.

Thus, $a_{1} a_{2}=2^{m_{1}+m_{2}} \cdot 3^{n_{1}+n_{2}}$, and $m_{1}+m_{2}, \quad n_{1}+n_{2}$ are both even, so $a_{1} a_{2}$ is a perfect square.
Therefore, the answer is $(C)$.