17. Among all the divisors of 2020, there are ( ) numbers that have more than 3 divisors (for example, 12 has six divisors, which are $1,2,3,4,6$ and 12).
(A) 6
(B) 7
(C) 8
(D) 9
(E) 10
17. B.

Given $2020=2^{2} \times 5 \times 101$, we know that 2020 has 12 divisors, which are $1, 2, 4, 5, 10, 20, 101, 202, 404, 505, 1010, 2020$. Among these, the divisors $1, 2, 4, 5, 101$ have no more than 3 divisors.
Therefore, the number of divisors that meet the requirement is $12-5=7$.