Example 6 What is the smallest positive integer that can be expressed 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 (AIME))


Solution: Let the required positive integer be $A$. According to the problem, $A$ can be expressed as (where $p$, $n$, $k$ are all integers)
$$
\begin{aligned}
A & =(p+1)+(p+2)+\cdots+(p+9) \\
& =9 p+45, \\
A & =(n+1)+(n+2)+\cdots+(n+10) \\
& =10 n+55, \\
A & =(k+1)+(k+2)+\cdots+(k+11) \\
& =11 k+66 .
\end{aligned}
$$

From (1), (2), and (3) we get
$$
\begin{array}{l}
9 p=10(n+1), \\
10 n=11(k+1) .
\end{array}
$$

From (4) and (5), we know that $n$ is a multiple of 11, and when divided by 9, the remainder is 8. Therefore, the smallest positive integer for $n$ is 44.
Thus, the smallest positive integer for $A$ $=10 \times 44+55=$
495.