Proving a palindromic integer with an even number of digits is divisible by 11 I'm in an introductory course for discrete math so I'm a novice at English proofs. I'm not sure if my reasoning here is valid or if I'm using modular arithmetic correctly. Specifically the line I marked with $(**)$. I would appreciate any feedback. Sorry if any part of the proof seems obvious we are generally expected to spell everything out.

Objective: Prove every palindromic integer with an even number of digits is divisible by $11$.
Proof:
Consider a palindromic integer $p$ in the form of $x_{1} x_{2} …. x_{n-1}x_{n}x_{n}x_{n-1}...x_{2}x_{1}$ where $p$ has $2n$ digits.
This can be expanded as:
$$x_{1} + x_{2}\cdot10 + … + x_{n}\cdot10^{n} + x_{n}\cdot10^{n+1} + … + x_{2}\cdot10^{2n} + x_{1}\cdot10^{2n+1}$$
$(**)$ $10 \equiv -1 \pmod{11}$ so if we take $\pmod{11}$ of the expression we can replace $10\equiv -1$.
$$x_{1} + x_{2}\cdot(-1) + … + x_{n}\cdot(-1)^{n} + x_{n}\cdot(-1)^{n+1} + … + x_{2}\cdot(-1)_{2n} + x_{1}\cdot(-1)^{2n+1}$$
Since we know $2n$ is even (and therefore $2n + 1$ is odd), and $(-1)^{a} = 1$ when $a$ is even and $= -1$ when $a$ is odd we can rewrite the expression as:
$$x_{1} - x_{2} + … + x_{n} - x_{n} + … + x_{2} - x_{1} = 0$$
Therefore since $p \equiv 0 \pmod{11}$, we have that $11$ divides $p$.

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*}
