6.77 Find the maximum value of $k$ such that $3^{11}$ can be expressed as the sum of $k$ consecutive positive integers.
[Solution] Let
$$3^{11}=(n+1)+(n+2)+\cdots+(n+k),$$

where $n$ is a non-negative integer, and $k$ is a natural number, i.e.,
$$\begin{array}{l}
3^{11}=n k+\frac{k(k+1)}{2} \\
2 \cdot 3^{11}=k(2 n+k+1)
\end{array}$$

Clearly, $k<2 n+k+1$. To make $k$ as large as possible, while $n$ is a non-negative integer, the best possible scenario is
$$k=2 \cdot 3^{5}, 2 n+k+1=3^{6},$$

In this case, $n=121, k=486$.
Therefore, the largest $k$ is 486.