2.44 Write down the two-digit numbers from 19 to 80 in succession. Ask: Is the resulting number
$$
19202122 \cdots 787980
$$

divisible by 1980?
(14th All-Union Mathematical Olympiad, 1980)
［Solution］Let $A=19202122 \cdots 7980$.
Obviously, $20 \mid A$.
Since $100^{k}=(99+1)^{k}=99 M+1$.
Where $M$ is a positive integer, $k$ is a positive integer.
Therefore,
$$
100^{k} \equiv 1
$$
$(\bmod 99)$.
Thus, we have
$$
\begin{aligned}
A & =19 \cdot 100^{61}+20 \cdot 100^{60}+\cdots+79 \cdot 100+80 \\
& \equiv(19+20+21+\cdots+79+80) \\
& \equiv 31 \cdot 99 \\
& \equiv 0 \quad(\bmod 99) .
\end{aligned}
$$

Therefore,
$99 \mid A$.
Since $(20,99)=1$, we have
$$
20 \cdot 99=1980 \mid A \text {. }
$$