The question is about the sum of a $5$-subset of $\{1,2,\dots,80\}$, taken modulo $10$.
Yes, there are some tricky ways to enumerate those.
Denote $\varepsilon_k = e^{\pi k i/5}$, $k=1,\dots,9$, $10$th roots of unity. Then
$$\tag{1}
\sum_{k=0}^9 \varepsilon_k^n = \begin{cases}
10, & 10\mid n,\\
0, & 10\nmid n.
\end{cases}
$$
Note that
$$
F(u,z) = \prod_{j = 1}^{80} (1+ u z^j) = \sum_{j,n} A(j,n) u^j z^n,
$$
where $A(j,n)$ is the number of $j$-subsets with sum $n$.
Therefore, the number of $5$-subsets with sum equal to $0$ modulo $10$ is equal, thanks to $(1)$, to
$$
[u^5]\frac{1}{10}\sum_{k=0}^9 F(u,\varepsilon_k),
$$
(the coefficient before $u^5$).
If $\varepsilon_k$ is a primitive root (i.e. $k=1,3,7,9$), then (noting that $\varepsilon^{10}_k = 1$)
$$
F(u,\varepsilon_k) = \Big(\prod_{n=0}^{9}(1+u \varepsilon_k^{n})\Big)^{8} = \big(1 - (-u)^{10}\big)^{8} = (1-u^{10})^8,
$$
since $\varepsilon_k^n$, $n=0,\dots,9$, are different $10$th roots of unity.
If $k=2,4,6,8$, then $\varepsilon_k$ is a primitive $5$th root of unity, so, similarly,
$$
F(u,\varepsilon_k) = \big(1 - (-u)^{5}\big)^{16} = (1+u^5)^{16}.
$$
Further,
$$
F(u,\varepsilon_5) = \big(1-(-u)^2\big)^{40} = (1-u^2)^{40}.
$$
Finally,
$$
F(u,\varepsilon_0) = F(u,1) (1+u)^{80}.
$$
Therefore, the number of $5$-subsets with sum divisible by $10$ is
$$
[u^5]\frac{1}{10}\big( 4(1-u^{10})^8 + 4(1+u^5)^{16} + (1-u^2)^{40} + (1+u)^{80} \big)\\
= \frac25 {16 \choose 1} + \frac1{10}{80\choose 5} = 2\,404\,008.
$$
The number of $4$-subsets:
$$
[u^4]\frac{1}{10}\big( 4(1-u^{10})^8 + 4(1+u^5)^{16} + (1-u^2)^{40} + (1+u)^{80} \big)\\
= \frac1{10} {40 \choose 2} + \frac1{10}{80\choose 4} = 158\,236.
$$
I leave you other last digits as an exercise.
If we were interested in the sum modulo some prime number, the computation would be much easier.
