The last one

$\begin{aligned} S & \equiv 1^2+2^2+3^2+\cdots + 2017^2 \text{ (mod 10)} \\ & = \sum_{n=1}^{2017} n^2 \text{ (mod 10)} \\ & = \frac {2017\times 2018 \times 4035}6 \text{ (mod 10)} \\ & = 2017 \times 1009 \times 1345 \text{ (mod 10)} & \small \color{#3D99F6} \text{Since 1345 ends with 5 and 2017 and 1009 are odd,} \\ & = \boxed{5} \text{ (mod 10)} \end{aligned}$

When squaring a number, the last digit of the result is going to be equal to the square of the last digit of the original number.

In general:

$1^2 => 1$

$2^2 => 4$

$3^2 => 9$

$4^2 => 6$

$5^2 => 5$

$6^2 => 6$

$7^2 => 9$

$8^2 => 4$

$9^2 => 1$

$0^2 => 0$

However, we know that, in every set of 10 digits, each of the above appears exactly once, so they appear the same number of times. Therefore, as $1^2$ appears the same number as $3^2$ , the sum of those two leads us to $0 \mod{10}$ .

The numbers pair up as follows to leave $0 \mod{10}$ : 1s and 3s 2s and 4s 6s and 8s 7s and 9s 0s

The only number which isn't complemented is 5. However, in every 20 integers, the 5 appears twice, cancelling itself out. Therefore, we have a conclusion that the sum of the squares of 20 consecutive integers $= 0 \mod{10}$ .

As 2020 is a multiple of 20, the sum of the squares up to 2020 equals $0 \mod{10}$ . However, this includes $2018^2 + 2019^2 + 2020^2$ . Subtracting $4 + 1 + 0$ , we get $5 \mod{10}$ . Such, the answer is $\boxed{5}$

I don't have an innovative solution to the problem. I simply did it by the formula of sum of first $n$ squares, that is $\dfrac{(n)(n+1)(2n+1) }{6}$ . After substituting $2017$ , we get the sum equivalent to $2017\times1345\times1009$ , hence the last digit= $\boxed{5}$

$\frac{n(n+1)(2n+1)}{6}$

When $n = 2017$

$= \frac{ 2017(2018)(2017 \cdot 2 + 1)}{6}$

$= \frac{(4070306)(4035)}{6}$

$= 2737280785$

Hence the last digit is $\boxed{5}$