11. Find the maximum value of $k$ such that $3^{11}$ can be expressed as the sum of $k$ consecutive positive integers.
i. Find the maximum positive integer $k$ such that $3^{11}=(n+1)+(n+2)+\cdots+(n+k)$ holds (where $n$ is a non-negative integer). From the right side, we have $=\frac{k(k+2 n+1)}{2}$ $->K \cdot(k+2 n+1)=2 \cdot 3^{11}$. To make the smaller factor $k$ as large as possible, $n$ must be non-negative $\Rightarrow k=2 \cdot 3^{5}=$ 486. (At this point, $n=121,3^{11}=122+123+\cdots+607$ )