$30 \cdot 15$ Let $a, b$ be integers, and form four propositions:
(1) If $a+5b$ is even, then $a-3b$ is also even.
(2) If $a+b$ is divisible by 3, then $a, b$ are both divisible by 3.
(3) If $a+b$ is a prime number, then $a-b$ is definitely not a prime number.
(4) If $c=a+b \neq 0$, then $\quad \frac{a^{3}-b^{3}}{a^{3}+c^{3}}=\frac{a-b}{a+c}$.

The number of correct propositions among the above is
(A) 1.
(B) 2.
(C) 3.
(D) 4.
(China "Zu Chongzhi Cup" Junior High School Mathematics Invitational Competition, 1989)
[Solution] Let $a+5b=2n$ (where $n=1,2,3, \cdots$), then $a=2n-5b$. Therefore, $a-3b=2n-8b=2(n-4b)$. Hence (1) holds.
Let $a+b=3n$ (where $n=1,2,3, \cdots$), then $n=\frac{a}{3}+\frac{b}{3}$.
When $n=1$, then $\frac{a}{3}<1, \frac{b}{3}<1$. Therefore, (2) does not hold.
If $a+b$ is a prime number, $a-b$ may also be a prime number, such as $14+3=17, 14-3=11$ are both prime numbers. Hence (3) does not hold.
Since $c=a+b \neq 0$, we have
\[
\begin{aligned}
\frac{a^3-b^3}{a^3+c^3} & =\frac{(a-b)(a^2+ab+b^2)}{(a+c)(a^2-ac+c^2)} \\
& =\frac{(a-b)(a^2+ab+b^2)}{(a+c)(a^2-ab-a^2+a^2+2ab+b^2)} \\
& =\frac{a-b}{a+c},
\end{aligned}
\]
Thus (4) holds.
Therefore, the answer is $(B)$.