Sum prime is fixed

The prime factorization of $2018$ is $2 \cdot 1009$ , so the distinct prime numbers are $2$ and $1009$ , and $2 + 1009 = \boxed{1011}$ .

$\text{Plus}\text{@@}\text{Table}[\text{Times}\text{@@}p,\{p,\text{FactorInteger}[2018]\}]$ is $1011$

My solution permits a prime factor to be present more than once, i.e., 2048 would yield an answer of 22.