$30 \cdot 13$ Choose five numbers from $1,2,3,4,5,6,7,8,9$. Then
（1）there must be two numbers that are coprime;
（2）there must be one number that is a multiple of another;
（3）there must be one number whose double is a multiple of another.
Among the above conclusions, the number of correct ones is
(A) 0.
(B) 1.
(C) 2.
(D) 3.
(Chinese Jiangsu Province Junior High School Mathematics Competition, 1988)
[Solution] Since five numbers are chosen from the nine consecutive natural numbers $1 \sim 9$, there must be two positive integers that are adjacent, otherwise it would include 1. Adjacent positive integers are always coprime, and 1 is coprime with any natural number, so (1) holds.
Choosing $5,6,7,8,9$ can negate (2).
Divide these nine natural numbers into three groups: $(1,2,4,8)$, $(3,6,9)$, and $(5,7)$.
In the first two groups, any two numbers chosen from each group will always have one number that is a multiple of the other's double.
Thus, if all five numbers are taken from the first two groups, conclusion (3) clearly holds;
and if four or three of the five numbers are taken from the first two groups, at least two numbers will be taken from one of the groups, so conclusion (3) holds.
Therefore, the answer is $(C)$.