Example 18 (Shanghai Mathematics Competition Question) Let the natural number $\overline{62 \alpha \beta 427}$ be a multiple of 99, find $\alpha, \beta$.
Since $62 \alpha \beta 427$ is a multiple of 99, it is both a multiple of 9 and 11.
Because $\overline{62 \alpha \beta 427}$ is a multiple of 9, then
$6+2+\alpha+\beta+4+2+7=9 m \quad(m \in \mathbf{N})$, which means $\alpha+\beta+3=9(m-2)$.
Since $0 \leqslant \alpha \leqslant 9,0 \leqslant \beta \leqslant 9$, then
$3 \leqslant \alpha+\beta+3 \leqslant 21$.
Thus, $\alpha+\beta+3=9$ or $\alpha+\beta+3=18$, which means $\alpha+\beta=6$ or $\alpha+\beta=15$.
Because $\overline{62 \alpha \beta 427}$ is a multiple of 11, then
$(6+\alpha+4+7)-(2+\beta+2)=11 k\left(k \in \overline{\mathbf{Z}^{-}}\right)$, which means $\alpha-\beta+13=11 k$.
Since $4 \leqslant \alpha-\beta+13 \leqslant 22$, then
$\alpha-\beta+13=11$ or $\alpha-\beta+13=22$, which means
$\alpha-\beta=-2$ or $\alpha-\beta=9$.
Thus, we can get four systems of equations
$$
\left\{\begin{array} { l } 
{ \alpha + \beta = 6 , } \\
{ \alpha - \beta = - 2 , }
\end{array} \quad \left\{\begin{array} { l } 
{ \alpha + \beta = 6 , } \\
{ \alpha - \beta = 9 , }
\end{array} \quad \left\{\begin{array} { l } 
{ \alpha + \beta = 1 5 , } \\
{ \alpha - \beta = - 2 , }
\end{array} \quad \left\{\begin{array}{l}
\alpha+\beta=15, \\
\alpha-\beta=9 .
\end{array}\right.\right.\right.\right.
$$

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 $\alpha=2, \beta=4$.