Let $x_0,x_1,x_2,\dotsc$ be a sequence of numbers, where each $x_k$ is either $0$ or $1$. For each positive integer $n$, define 
\[S_n = \sum_{k=0}^{n-1} x_k 2^k\]
Suppose $7S_n \equiv 1 \pmod{2^n}$ for all $n \geq 1$. What is the value of the sum  
\[x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022}?\]
$\textbf{(A) } 6 \qquad \textbf{(B) } 7 \qquad \textbf{(C) }12\qquad \textbf{(D) } 14\qquad \textbf{(E) }15$
In binary numbers, we have \[S_n = (x_{n-1} x_{n-2} x_{n-3} x_{n-4} \ldots x_{2} x_{1} x_{0})_2.\]
It follows that \[8S_n = (x_{n-1} x_{n-2} x_{n-3} x_{n-4} \ldots x_{2} x_{1} x_{0}000)_2.\]
We obtain $7S_n$ by subtracting the equations:
\[\begin{array}{clccrccccccr}   & (x_{n-1} & x_{n-2} & x_{n-3} & x_{n-4} & \ldots & x_2 & x_1 & x_0 & 0 & 0 & 0 \ )_2 \\ -\quad & & & & (x_{n-1} & \ldots & x_5 & x_4 & x_3 & x_2 & x_1 & x_0)_2 \\ \hline   & & & & & & & & & & &  \\ [-2.5ex]   & ( \ \ ?& ? & ? & 0 \ \ \ & \ldots & 0 & 0 & 0 & 0 & 0 & 1 \ )_2 \\ \end{array}\]
We work from right to left:
\begin{alignat*}{6} x_0=x_1=x_2=1  \quad &\implies \quad &x_3 &= 0& \\  \quad &\implies \quad &x_4 &= 1& \\  \quad &\implies \quad &x_5 &= 1& \\ \quad &\implies \quad &x_6 &= 0& \\  \quad &\implies \quad &x_7 &= 1& \\  \quad &\implies \quad &x_8 &= 1& \\ \quad &\quad \vdots & & & \end{alignat*}
For all $n\geq3,$ we conclude that

$x_n=0$ if and only if $n\equiv 0\pmod{3}.$
$x_n=1$ if and only if $n\not\equiv 0\pmod{3}.$
Finally, we get $(x_{2019},x_{2020},x_{2021},x_{2022})=(0,1,1,0),$ from which \[x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022} = \boxed{\textbf{(A) } 6}.\]
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
~MRENTHUSIASM