The number 2018 has exactly two divisors that are prime numbers. The sum of these two prime numbers is
(A) 793
(B) 1011
(C) 38
(D) 209
(E) 507
Since $2018=2 \times 1009$, and both 2 and 1009 are prime numbers, then the required sum is $2+1009=1011$.

Note: In the question, we are given that 2018 has exactly two divisors that are prime numbers, and since 2 is a prime divisor of 2018, then 1009 must be the other prime divisor.

ANSWER: (B)