Question

设 $x_0,x_1,x_2,\dotsc$ 是一个数列，其中每个 $x_k$ 都是 $0$ 或 $1$ 。对每个正整数 $n$ ，定义。假设对所有 $n \geq 1$ ，都有 $7S_n \equiv 1 \pmod{2^n}$ 。求的值。 $S_n = \sum_{k=0}^{n-1} x_k 2^k$ $x_{2019} + 2x_{2020} +$ $4x_{2021} + 8x_{2022}?$

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 $x_{2019} + 2x_{2020} +$ $4x_{2021} + 8x_{2022}?$

$6$

$7$

$12$

$14$

$15$

答案：A

知识点：模运算进制 2的幂

难度评级：2390

解答：

首先注意，因此只需求出 $0\le S_n<2^n$ 。另外， $\frac{S_{2023}-S_{2019}}{2^{2019}}.$

由 $m_n\in\{0,1,\ldots,6\}$ ，可取某个整数 $m_n2^n\equiv-1\pmod7$ ，使。又由，有因而。接下来只需寻找使能被 $7$ 整除的，即满足的整数。 $7S_n=m_n2^n+1.$

当 $2^3\equiv1\pmod7$ 时，所以 $2019\equiv0\pmod3$ 因此 $m_{2019}=6$ 。 $2023\equiv1\pmod3$ ， $2m_{2023}\equiv-1\pmod7$ ， $m_{2023}=3$ 。 $\begin{gathered}S_{2019}=\frac{6\cdot2^{2019}+1}{7},\\ S_{2023}=\frac{3\cdot2^{2023}+1}{7}.\end{gathered}$

当时，所以因此 $\frac{S_{2023}-S_{2019}}{2^{2019}} =\frac{3\cdot2^4-6}{7}=6.$

所以正确答案是 A。

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 .

2022 AMC 10B 第 25 题

答案：A

难度评级：2390

解答：

其他年份的第 25 题