# 2022 AMC 10B Problems/Problem 25

The following problem is from both the 2022 AMC 10B #25 and 2022 AMC 12B #23, so both problems redirect to this page.

## Problem

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 $n \geqslant 1$. What is the value of the sum $$x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022}?$$ $\textbf{(A) } 6 \qquad \textbf{(B) } 7 \qquad \textbf{(C) }12\qquad \textbf{(D) } 14\qquad \textbf{(E) }15$

## Solution 1

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: $$\begin{array}{clccrccccccr} & (x_{n-1} & x_{n-2} & x_{n-3} & x_{n-4} & \ldots & x_2 & x_1 & x_0 & 0 & 0 & 0 \ )_2 \\ -\quad & & & & (x_{n-1} & \ldots & x_5 & x_4 & x_3 & x_1 & x_1 & x_0)_2 \\ \hline & & & & & & & & & & & \\ [-2.5ex] & ( \ \ ?& ? & ? & 0 \ \ \ & \ldots & 0 & 0 & 0 & 0 & 0 & 1 \ )_2 \\ \end{array}$$ 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

## Solution 2

First, notice that $$x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022} = \frac{S_{2023} - S_{2019}}{2^{2019}}.$$ Then since $S_n$ is the modular inverse of $7$ in $\mathbb{Z}_{2^n}$, we can perform the Euclidean Algorithm to find it for $n = 2019,2023$.

Starting with $2019$, \begin{align*} 7S_{2019} &\equiv 1 \pmod{2^{2019}} \\ 7S_{2019} &= 2^{2019}k + 1. \end{align*} Now, take both sides $\operatorname{mod} \ 7$: $$0 \equiv 2^{2019}k + 1 \pmod{7}.$$ Using Fermat's Little Theorem, $$2^{2019} = (2^{336})^6 \cdot 2^3 \equiv 2^3 \equiv 1 \pmod{7}.$$ Thus, $$0 \equiv k + 1 \pmod{7} \implies k \equiv 6 \pmod{7} \implies k = 7j + 6.$$ Therefore, $$7S_{2019} = 2^{2019} (7j + 6) + 1 \implies S_{2019} = \frac{2^{2019} (7j + 6) + 1}{7}.$$

We may repeat this same calculation with $S_{2023}$ to yield $$S_{2023} = \frac{2^{2023} (7h + 3) + 1}{7}.$$ Now, we notice that $S_n$ is basically an integer expressed in binary form with $n$ bits. This gives rise to a simple inequality, $$0 \leqslant S_n \leqslant 2^n.$$ Since the maximum possible number that can be generated with $n$ bits is $$\underbrace{{11111\dotsc1}_2}_{n} = \sum_{k=0}^{n-1} 2^k = 2^n - 1 \leqslant 2^n.$$ Looking at our calculations for $S_{2019}$ and $S_{2023}$, we see that the only valid integers that satisfy that constraint are $j = h = 0$. $$\frac{S_{2023} - S_{2019}}{2^{2019}} = \frac{\tfrac{2^{2023} \cdot 3 + 1}{7} - \tfrac{2^{2019} \cdot 6 + 1}{7}}{2^{2019}} = \frac{2^4 \cdot 3 - 6}{7} = \boxed{\textbf{(A) } 6}.$$ ~ $\color{magenta} zoomanTV$

## Solution 3

As in Solution 2, we note that $$x_{2019}+2x_{2020}+4x_{2021}+8x_{2022}=\frac{S_{2023}-S_{2019}}{2^{2019}}.$$ We also know that $7S_{2023} \equiv 1 \pmod{2^{2023}}$ and $7S_{2019} \equiv 1 \pmod{2^{2019}}$, this implies: $$\textbf{(1) } 7S_{2023}=2^{2023}\cdot{x} + 1,$$ $$\textbf{(2) } 7S_{2019}=2^{2019}\cdot{y} + 1.$$ Dividing by $7$, we can isolate the previous sums: $$\textbf{(3) } S_{2023}=\frac{2^{2023}\cdot{x} + 1}{7},$$ $$\textbf{(4) } S_{2019}=\frac{2^{2019}\cdot{y} + 1}{7}.$$ The maximum value of $S_n$ occurs when every $x_i$ is equal to $1$. Even when this happens, the value of $S_n$ is less than $2^n$. Therefore, we can construct the following inequalities: $$\textbf{(3) } S_{2023}=\frac{2^{2023}\cdot{x} + 1}{7} < 2^{2023},$$ $$\textbf{(4) } S_{2019}=\frac{2^{2019}\cdot{y} + 1}{7} < 2^{2019}.$$ From these two equations, we can deduce that both $x$ and $y$ are less than $7$.

Reducing $\textbf{1}$ and $\textbf{2}$ $\pmod{7},$ we see that $$2^{2023}\cdot{x}\equiv 6\pmod{7},$$ and $$2^{2019}\cdot{y}\equiv 6\pmod{7}.$$

The powers of $2$ repeat every $3, \pmod{7}.$

Therefore, $2^{2023}\equiv 2 \pmod 7$ and $2^{2019} \equiv 1 \pmod {7}.$ Substituing this back into the above equations, $$2x\equiv{6}\pmod{7}$$ and $$y\equiv{6}\pmod{7}.$$

Since $x$ and $y$ are integers less than $7$, the only values of $x$ and $y$ are $3$ and $6$ respectively.

The requested sum is \begin{align*} \frac{S_{2023}-S_{2019}}{2^{2019}} &= \frac{\frac{2^{2023}\cdot{x} + 1}{7} - \frac{2^{2019}\cdot{y} + 1}{7}}{2^{2019}} \\ &= \frac{1}{2^{2019}}\left(\frac{2^{2023}\cdot{3} + 1}{7} -\left(\frac{2^{2019}\cdot{6} + 1}{7} \right)\right) \\ &= \frac{3\cdot{2^4}-6}{7} \\ &= \boxed{\textbf{(A) } 6}. \end{align*} -Benedict T (countmath1)

## Video Solutions

~ ThePuzzlr

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

~ pi_is_3.14

## Video Solution by The Power of Logic

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