30. Given $\mathrm{x}, \mathrm{y}$ are distinct prime numbers, if $\frac{621 x y}{x+y}$ is an integer, then the number of ordered pairs $(\mathrm{x}, \mathrm{y})$ is $(\quad)$
(A) 8
(B) 6
(C) 4
(D) 2
Answer: (B)

Solution: Since $\mathrm{x}, \mathrm{y}$ are distinct prime numbers, $(\mathrm{xy}, \mathrm{x}+\mathrm{y})=1$, therefore, $\mathrm{x}+\mathrm{y}$ is a factor of 621.

Notice that $621=3^{3} \times 23$, which has 8 factors: $1,3,9,23,27,69,207,621$, all of which are odd,

so $\mathrm{x}+\mathrm{y}$ should be odd, and one of $\mathrm{x}$ and $\mathrm{y}$ must be 2, the other being a prime number greater than 2,

thus the pairs $(x, y)$ that meet the requirements are: $(2,7),(7,2),(2,67),(67,2),(2,619),(619,2)$.

Therefore, the answer is (B).