3. Find the non-zero natural numbers $n \in \mathbb{N}^{*}$, for which the sum $S=2^{n}+4^{n}+6^{n}+\ldots+1888^{n}$ is divisible by 10.
Solution.

$$
\begin{aligned}
& U\left(2^{n}\right)= \begin{cases}6, & n=4 k \quad k \in \mathbb{N}^{*} \\
2, & n=4 k+1 \\
4, & n=4 k+2 \\
8, & n=4 k+3\end{cases} \\
& U\left(4^{n}\right)= \begin{cases}6, & n=2 k \\
4, & n=2 k+1 \quad k \in \mathbb{N}^{*} \\
(1p)\end{cases} \\
& U\left(6^{n}\right)=6, \quad k \in \mathbb{N}^{*} \quad(1p) \\
& U\left(8^{n}\right)= \begin{cases}6, & n=4 k \quad k \in \mathbb{N}^{*} \\
8, & n=4 k+1 \\
4, & n=4 k+2 \\
2, & n=4 k+3\end{cases}
\end{aligned}
$$

Therefore,

$$
U(S)=U\left(2^{n}+4^{n}+6^{n}+8^{n}\right)= \begin{cases}4, & n=4 k \quad k \in \mathbb{N}^{*} \\ 0, & n=4 k+1 \\ 0, & n=4 k+2 \\ 0, & n=4 k=3\end{cases}
$$

The numbers sought are of the form $n=4 k, k \in \mathbb{N}^{*}$.