Two digits of 2018

As $81 = 3^4$ it is good form to see first when is $n^3 - 9n + 27$ divisible by 3 as prime congruences are easy to solve. By applying Fermat's theorem we see that $n^3 - 9n + 27 \equiv 0 \mod 3$ if and only if $n \equiv 0 \mod 3$ . Therefore if a solution exists it will have the form $n = 3k$ . Let's plug this into the equation to get $(3k)^3 - 9 (3k) + 27 = 27k^3 - 27k + 27 = 27(k^3 - k + 1)$ .

If $27(k^3 - k + 1) = 81q$ then $k^3 - k + 1 = 3q$ and therefore $k^3 - k + 1 \equiv 0 \mod 3$ but then this would imply $1 \equiv 0 \mod 3$ . Therefore, if we assume a solution exists then we get a contradiction. So there must be no solutions.