Remainder Properties #2

Since $5$ is a prime, Euler's theorem applies and Euler's totient function $\phi(5)=5-1 = 4$ . Then we have:

$\begin{aligned} \sum_{n=1}^9 n^{2016+n} & \equiv \sum_{n=1}^9 n^{2016+n \bmod \phi (5)} \equiv \sum_{n=1}^9 n^{2016+n \bmod 4} \equiv \sum_{n=1}^9 n^{n \bmod 4} \pmod 5 \\ & = 1^1 + 2^2 + 3^3 + 4^0 + 5^1 + 6^2 + 7^3 + 8^0 + 9^1 \pmod 5 \\ & = 1 + 4 + 2 + 1 + 0 + 1^2 + 2^3 + 1 + 4 \pmod 5 \\ & = 1 + 4 +2 + 1 + 0 + 1 + 3 + 1 + 4 \pmod 5 \\ & = \boxed 2 \pmod 5 \end{aligned}$

Answer
= { 1^2017 + 2^2018 + 3^2019 + 4^2020 + 5^2021 + 6^2022 + 7^2023 + 8^2024 + 9^2025 } mod 5
= { 1^(2017 mod 4) + 2^(2018 mod 4) + 3^(2019 mod 4) + 4^(2020 mod 4) + 5^(2021 mod 4) + 6^(2022 mod 4) + 7^(2023 mod 4) + 8^(2024 mod 4) + 9^(2025 mod 4) } mod 5
= { 1^1 + 2^2 + 3^3 + 4^0 + 5^1 + 6^2 + 7^3 + 8^0 + 9^1 } mod 5
= { 1 + 4 + 2 + 1 + 0 + 1 + 3 + 1 + 4 } mod 5
= 17 mod 5
= 2