2022 AMC 10B Problem 25

The desired sum is $$\frac{S_{2023}-S_{2019}}{2^{2019}}.$$ Also, $0\le S_n<2^n.$ Therefore, for a unique $m_n\in\{0,1,\ldots,6\},$ $$7S_n=m_n2^n+1.$$ Reducing modulo $7$ gives $m_n2^n\equiv-1\pmod7.$ Since $2^3\equiv1\pmod7$ and $2019\equiv0\pmod3,$ we get $m_{2019}=6.$ Since $2023\equiv1\pmod3,$ we have $2m_{2023}\equiv-1\pmod7,$ so $m_{2023}=3.$ Hence $$\begin{gathered}S_{2019}=\frac{6\cdot2^{2019}+1}{7},\\ S_{2023}=\frac{3\cdot2^{2023}+1}{7}.\end{gathered}$$ Finally, $$\frac{S_{2023}-S_{2019}}{2^{2019}} =\frac{3\cdot2^4-6}{7}=6.$$ Thus, the correct answer is A .