14. Can $1971^{26}+1972^{27}+1973^{28}$ be divisible by 3?
14. $1971 \equiv 0(\bmod 3)$, so, $1971^{26} \equiv 0(\bmod 3)$.

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

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

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