Congruence proof involving $x$ and a summation of its digits.

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Let $(x_n, x_{n-1},x_{n-2},...,x_0)_{10}$ be the representation of $x$ in base $10$. I was requested to prove that

$x \equiv x_0-x_1+x_2-x_3+...+(-1)^nx_n \mod{(11)}$.

I started by noticing that $11 \equiv 1 \mod 11 \implies 11^k \equiv 1 \mod 11 \implies z11^k \equiv z \mod 11$, for $k>0$. I supposed one of these three facts would turn to be useful, but I hardly found a use for them in my attempts.

I know what I essentially have to prove is that

$$x-(x_0-x_1+x_2-x_3+...+(-1)^nx_n) = x - \sum_{0}^n (-1)^nx_n =11m$$

(i.e., it's divisible by $11$). I have not, though, found a way to prove this at all. How would one go about proving these?

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Since $10\equiv -1\mod{11},\;\;$ we get that

$10^k\equiv (-1)^k\mod{11}\;\;$ for all $\;\;k\in\mathbb{N}$.

Moreover

$x=x_0+10x_1+10^2x_2+10^3x_3+\ldots+10^nx_n$,

therefore

$x\equiv x_0+(-1)x_1+(-1)^2x_2+(-1)^3x_3+\ldots+(-1)^nx_n \mod{11}$

that is

$x\equiv x_0-x_1+x_2-x_3+\ldots+(-1)^nx_n \mod{11}$.