Number theory

Let $(x+1)+(x+2)+.....+(x+k)=3^{11}.$
Then $\frac{k}{2}(2x+k+1)=3^{11};i.e.,k(2x+k+1)=2\times 3^{11}.$ Let $k=2^{\alpha}3^{\beta}.$ Then $k^2< (2\times 3^{11}) \Rightarrow 2^{2\alpha}3^{2\beta})<(2\times 3^{11})\Rightarrow 3^{2\beta}<(2\times 3^{11})<3^{12}\Rightarrow 2\beta<12\Rightarrow \beta\leq 5.$ Since $\alpha\leq 1,$ we have $k\leq (2\times 3^5)=486;k$ takes this value when $x=121;$
Therefore $\boxed{486}$ is the largest possible value of $k.$