Power Divisibilities

$18^{2017} + 18^{2018} = 1 \times 18^{2017} + 18 \times 18^{2017} = 19 \times 18^{2017}$ . It is true that $28 = 7 \times 4$ . But $7$ is a factor of neither $18$ nor $19$ . Thus $28$ cannot be a factor of $18^{2017} + 18^{2018}$ . So our answer is $\boxed{28}$ .

Let $N = 18^{2017}+18^{2018} = 18^{2017}(1+18) = 2^{2017}\cdot 3^{4034} \cdot 19$ . Since $28 =2^2\cdot 7$ has a prime factor 7 which is not a factor of $N$ , $\boxed{28}$ is not a factor of $N$ . The rest of the options have powers of 2, 3, and 19 as factors which are the same or smaller than those of $N$ therefore, they are factors of $N$ .