$12 \cdot 63$ can be expressed as the smallest positive integer that can be represented as the sum of 9 consecutive integers, the sum of 10 consecutive integers, and the sum of 11 consecutive integers. 
(11th American Invitational Mathematics Examination, 1993)
\[ \text{Solution: Let } a \text{ be expressible as the sum of 9, 10, and 11 consecutive integers, then} \]
\[ \begin{array}{l}
a=l+(l+1)+\cdots+(l+8)=9 l+36, \\
a=m+(m+1)+\cdots+(m+9)=10 m+45, \\
a=n+(n+1)+\cdots+(n+10)=11 n+55,
\end{array} \]
\[ \text{where } l, m, n \in \mathbb{N}. \text{ Then} \]
\[ \left\{\begin{array}{l}
9 l+36=11 n+55, \\
10 m+45=11 n+55 .
\end{array}\right. \]
\[ \text{So} \]
\[ \begin{array}{c}
9 l=11 n+19, \quad 10 m=11 n+10 . \\
\left\{\begin{array}{l}
l=n+2+\frac{2 n+1}{9}, \\
m=n+1+\frac{n}{10} .
\end{array}\right.
\end{array} \]
\[ \text{Thus} \]
\[ \left\{\begin{aligned}
2 n+1 & \equiv 0 \quad(\bmod 9), \\
n & \equiv 0 \quad(\bmod 10) .
\end{aligned}\right. \]
\[ \text{The smallest } n \text{ satisfying the above is } n=40. \]
\[ \text{Therefore, the minimum value of } a \text{ is} \]
\[ 40+41+\cdots+50=495 . \]