imc

Verify that $a^3\equiv a(\mathrm{mod}6)$ manually for all $a\in \mathbb{Z}/6\mathbb{Z}$. We check: $0^3\equiv 0(\mathrm{mod}6)$, $1^3\equiv 1(\mathrm{mod}6)$, $2^3\equiv 8\equiv 2(\mathrm{mod}6)$, $3^3\equiv 27\equiv 3(\mathrm{mod}6)$, $4^3\equiv 64\equiv 4(\mathrm{mod}6)$, and $5^3\equiv 125\equiv 5(\mathrm{mod}6)$. We conclude that $a^3\equiv a(\mathrm{mod}6)$. Therefore, $$a_1+a_2+\cdots +a_{2018}\equiv a_1^3+a_2^3+\cdots +a_{2018}^3(\mathrm{mod}6).$$ Thus the answer is congruent to $2018^{2018}\equiv 2^{2018}(\mathrm{mod}6)=\boxed{4}$ because $2^n(\mathrm{mod}6)$ alternates with $2$ and $4$ when $n$ increases.