$2 \cdot 81971^{26}+1972^{27}+1973^{28}$ Can it be divisible by 3?
(Kiev Mathematical Olympiad, 1971)
[Solution] $1971 \equiv 0 \quad(\bmod 3)$, so
$$
1971^{26} \equiv 0 \quad(\bmod 3) .
$$

Also, $1972 \equiv 1(\bmod 3)$, so
$$
\begin{aligned}
1972^{27} & \equiv 1^{27} \quad(\bmod 3), \\
1973 & \equiv 2 \quad(\bmod 3), \\
1973^{28} & \equiv 2^{28} \quad(\bmod 3) .
\end{aligned}
$$

Thus, we have
$$
1971^{26}+1972^{27}+1973^{28} \equiv 2^{28}+1 \quad(\bmod 3) .
$$

Also, $2^{28}=4^{14} \equiv 1(\bmod 3)$, so
$$
2^{28}+1 \equiv 2 \quad(\bmod 3) .
$$

Therefore, $1971^{26}+1972^{27}+1973^{28}$ cannot be divisible by 3.