$9 \cdot 21$ Let the natural number $62 \alpha \beta 427$ be a multiple of 99, find $\alpha, \beta$.
(Shanghai Mathematical Competition, 1957)
[Solution] Since $62 \alpha \beta 427$ is a multiple of 99, it is both a multiple of 9 and 11.
Since $\overline{62 \alpha \beta 427}$ is a multiple of 9, then
$$
\begin{array}{c}
6+2+\alpha+\beta+4+2+7=9 m \quad(m \in N) . \\
\alpha+\beta+3=9(m-2) .
\end{array}
$$

Since
$$
0 \leqslant \alpha \leqslant 9, \quad 0 \leqslant \beta \leqslant 9,
$$

then
$$
3 \leqslant \alpha+\beta+3 \leqslant 21 \text {. }
$$

Thus
$$
\alpha+\beta+3=9 \text { or } \alpha+\beta+3=18 \text {. }
$$

That is
$$
\alpha+\beta=6 \text { or } \alpha+\beta=15 \text {. }
$$

Since $\overline{62 \alpha \beta 427}$ is a multiple of 11, then
$$
\begin{array}{c}
(6+\alpha+4+7)-(2+\beta+2)=11 k \quad\left(k \in \overline{Z^{-}}\right), \\
\alpha-\beta+13=11 k .
\end{array}
$$

Since
$$
4 \leqslant \alpha-\beta+13 \leqslant 22,
$$

then
$$
\alpha-\beta+13=11 \text { or } \alpha-\beta+13=22 \text {, }
$$

That is
$$
\alpha-\beta=-2 \text { or } \alpha-\beta=9 \text {. }
$$

Thus, we can obtain four systems of equations
$$
\begin{array}{l}
\left\{\begin{array} { l } 
{ \alpha + \beta = 6 , } \\
{ \alpha - \beta = - 2 , }
\end{array} \quad \left\{\begin{array}{l}
\alpha+\beta=6, \\
\alpha-\beta=9,
\end{array}\right.\right. \\
\left\{\begin{array} { l } 
{ \alpha + \beta = 1 5 , } \\
{ \alpha - \beta = - 2 , }
\end{array} \quad \left\{\begin{array}{l}
\alpha+\beta=15, \\
\alpha-\beta=9 .
\end{array}\right.\right.
\end{array}
$$

Solving the above systems of equations, since the sum and difference of two numbers have the same parity, and $0 \leqslant \alpha \leqslant 9, 0 \leqslant \beta \leqslant 9$, the only solution is $\alpha=2, \beta=4$.