Index game

Solution 1 :

$2016^{27}+2017^{27}+2018^{27}+2019^{27} \ (\text{mod} 4035) \\ \equiv (-2019)^{27}+(-2018)^{27}+2018^{27}+2019^{27} \ (\text{mod} 4035) \\ \equiv 0 \ (\text{mod} 4035)$

$\therefore$ Yes

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Solution 2 :

Note that $a^3+b^3 = (a+b)(a^2-ab+b^2)$

$2016^{27}+2017^{27}+2018^{27}+2019^{27} \\ = (2016^{27}+2019^{27})+(2017^{27}+2018^{27}) \\ = (2016^9+2019^9)(\cdots) + (2017^9+2018^9)(\cdots) \\ = (2016^3+2019^3)(\cdots) + (2017^3+2018^3)(\cdots) \\ = (2016+2019)(\cdots) + (2017+2018)(\cdots) \\ = 4035(\cdots) +4035(\cdots) \\ = 4035(\cdots)$

$\therefore$ Yes