Last digit

$S = U(~1^5 + 2^5 + 3^5 + \ldots + 99^5 ~)\\ S = U(~U(1^5 + 99^5) + U(2^5 + 98^5) + U(3^5 + 97^5) + U(\ldots) + U(49^5 + 51^5) + U(50^5)~) \\ S = U(0 + 0 + 0 + \ldots + 0 + 0) \\ S = 0$

The big $U$ represents function which returns unit digit of given integer.

For $1^5+2^5+3^5+\cdots+99^5$ , the last digit of each term repeats itself 10 times throughout the sequence so $(1^5+2^5+3^5+\cdots+99^5) \pmod{10}$ becomes $10(1^5+2^5+3^5+\cdots+9^5) \pmod{10}$ . Since $10\mod(10)\equiv{0}\mod(10)$ the product becomes $0(1^5+2^5+3^5+\cdots+9^5)\pmod {10}$ which is simply $\boxed{0}$

$\begin{aligned} S & = \sum_{k=1}^{99} k^5 \quad \quad \small \color{#3D99F6}{\text{Using the identity }\sum_{k=a}^b f(k) =\sum_{k=a}^b f(a+b-k)} \\ & = \frac 12 \sum_{k=1}^{99} \left(k^5 + (100-k)^5\right) \\ & = \frac 12 \sum_{k=1}^{99} \left(\color{#D61F06}{k^5} + (100^5-5\cdot 100^4\cdot k+10\cdot 100^3 \cdot k^2-10\cdot 100^2\cdot k^3+3\cdot 100\cdot k^4-\color{#D61F06}{k^5}) \right) \\ & = \frac 12 \sum_{k=1}^{99} \left(100^5-5\cdot 100^4\cdot k+10\cdot 100^3 \cdot k^2-10\cdot 100^2\cdot k^3+3\cdot 100\cdot k^4 \right) \\ & = 50 \sum_{k=1}^{99} \left(100^4-5\cdot 100^3\cdot k+10\cdot 100^2 \cdot k^2-10\cdot 100\cdot k^3+3k^4 \right) \end{aligned}$

We note that $S$ is divisible by $10$ , therefore the last digit of $S$ is $\boxed{0}$ .

Here we know that $(a+b)^p = 0$ $mod (a+b)$ or $(a+b)^p$ is divisible by $(a+b)$ hence $1^5 +99^5 + 2^5 + 98^5 + ........$ ( Like this pair wise) Hence we know $1^5+99^5$ is divisible by $(1+99) = 100$ and $(2^5+98^5)$ is divisible by $(2+98 )$ = $100$ and so on. Hence All have common factor as 100 . Hence the last digit must be 0.