30. 25 Among eight consecutive natural numbers, there are $k$ numbers that are pairwise coprime. The maximum value of $k$ is 
(A) 3.
(B) 4.
(C) 5.
(D) 6.
(Hubei Huanggang Region Junior High School Mathematics Competition, 1993)
［Solution］Let the eight natural numbers be:
$$
a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}, a_{7}, a_{8},
$$

and $a_{1}$ is an even number of the form $3k+1$, and $4 \mid a_{1}$.
Considering the Sieve of Eratosthenes, crossing out every other number, then $a_{3}, a_{5}, a_{7}$ are crossed out.
Since $a_{1}$ is of the form $3k+1$, $a_{3}$ is of the form $3k$, but it has already been crossed out, $a_{6}$ can serve as a $3k$ type number.
And since $a_{1}$ is even, and $4 \mid a_{1}$, then $4 \mid a_{5}$, but $a_{5}$ has already been crossed out.
$a_{2}$ is a $5r$ type number, so at most $a_{1}, a_{2}, a_{4}, a_{6}, a_{8}$ these five numbers are pairwise coprime.
Notice that if $a_{1}$ is not a $3k+1$ type even number, the sieve method will cross out more than 5 numbers.
Consider the numbers $4,5,6,7,8,9,10,11$ where $4,5,7,9,11$ are pairwise coprime.
Therefore, the answer is $(C)$.