Let $a_1,a_2,\dots,a_{2018}$ be a strictly increasing sequence of positive integers such that \[a_1+a_2+\cdots+a_{2018}=2018^{2018}.\]
What is the remainder when $a_1^3+a_2^3+\cdots+a_{2018}^3$ is divided by $6$?
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4$
Note that, because of Fermat's Little Theorem, for any integer $a$ relatively prime to 6, we have $a^2 \equiv 1 \pmod{6} \implies a^3 \equiv a \pmod{6}$. This is also trivially true if $a$ is not relatively prime to 6 because $a \equiv 0 \pmod{6}$. Therefore,
\[a_1+a_2+\cdots+a_{2018} \equiv a_1^3+a_2^3+\cdots+a_{2018}^3 \pmod{6}.\]
Therefore the answer is congruent to $2018^{2018}\equiv 2^{2018} \pmod{6} = \boxed{ \text{(E)}4}$ because $2^n \pmod{6}$ alternates with $2$ and $4$ when $n$ increases.
~Dolphindesigner