Proving a palindromic integer with an even number of digits is divisible by 11

You have some off-by-one errors, but you have the right idea. Note that: \begin{align*} p &= (x_1 + x_2 10 + \cdots + x_n 10^{n-1}) + (x_{n} 10^n + x_{n-1} 10^{n+1} + \cdots + x_1 10^{2n - 1}) \\ &= \sum_{k=1}^n x_k 10^{k-1} + \sum_{k=1}^n x_k 10^{2n-k} \\ &= \sum_{k=1}^n x_k10^{k-1}(1 + 10^{2n - 2k - 1}) \\ &\equiv \sum_{k=1}^n x_k10^{k-1}(1 + (-1)^{2n - 2k - 1}) \pmod {11} \qquad\qquad\text{since $10 \equiv -1 \pmod{11}$} \\ &\equiv \sum_{k=1}^n x_k10^{k-1}(1 -1) \pmod {11} ~~~~~~~~~~~~~~~~~~~\qquad\qquad\text{since $2n - 2k - 1$ is odd} \\ &\equiv 0 \pmod {11} \end{align*}