Sum till 89

$\begin{aligned} \sum_{i=1}^{89} i^{33} & \equiv \sum_{j=0}^8 \sum_{k=0}^9 (10j+k)^{33} \text{ (mod 10)} \\ & \equiv 9 \sum_{k=1}^9 k^{33} \text{ (mod 10)} \\ & \equiv (10-1)\left(\sum_{k=1}^4 \left(\blue{k^{33}+(10-k)^{33}}\right) + 5^{33} \right) \text{ (mod 10)} & \small \blue{\text{As }a^{33}+b^{33} = (a+b)\left(a^{32}-a^{31}b+a^{30}b^2 - \cdots + b^{32}\right)} \\ & \equiv -\left(\sum_{k=1}^4 \blue{10\sum_{n=0}^{32}k^{32-n}(10-k)^n} + 5^{33} \right) \text{ (mod 10)} \\ & \equiv - 0 - 5 \equiv \boxed 5 \text{ (mod 10)} \end{aligned}$

Since $\phi(10)=4$ and $33=4.8+1$ .we have ,the sum= $\sum_{i=1}^{89} i =4005 =5\mod10$ .

Note that $n^{33} \equiv n \mod 10$ .

Hence $\displaystyle \sum_{i=1}^{89} i^{33} \equiv \sum_{i=1}^{89} i\mod 10 \equiv \dfrac{89 \times 90}{2} \mod 10= \boxed{5}$ .