$32 \cdot 25$ Among the 101 integers $1,2,3, \cdots, 101$, the numbers that can be divided by both 3 and 5 and leave a remainder of 1 are
(A) 6.
(B) 7.
(C) 8.
(D) 9.
(2nd "Jinyun Cup" Junior High School Mathematics Invitational, 1985)
〔Solution〕 Obviously, numbers that leave a remainder of 1 when divided by both 3 and 5 (note that 3 and 5 are both prime numbers) can be expressed as
$$
3 \cdot 5 k+1=15 k+1 \quad(k=0,1,2 \cdots),
$$

Thus, among the first 101 natural numbers, there are $1,16,31,46,61,76,91$, a total of 7 numbers that meet the condition. Therefore, the answer is $(B)$.