2022 AMC 12B Problems/Problem 23

The following problem is from both the 2022 AMC 12B #23 and 2022 AMC 10B #25, 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 $n \geq 1$ . What is the value of the sum $x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022}$

$\textbf{(A)}~6\qquad\textbf{(B)}~7\qquad\textbf{(C)}~12\qquad\textbf{(D)}~14\qquad\textbf{(E)}~15\qquad$

Solution

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 $2019$ , $7S_{2019} \equiv 1 \pmod{2^{2019}}$ $7S_{2019} = 2^{2019}k + 1$ Now, take both sides $\text{mod } 7$ $0 \equiv 2^{2019}k + 1 \pmod{7}$ 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 2 (Base-2 Analysis)

We solve this problem with base 2. To avoid any confusion, for a base-2 number, we index the $k$ th rightmost digit as digit $k-1$ .

We have $S_n = \left( x_{n-1} x_{n-2} \cdots x_1 x_0 \right)_2$ .

In the base-2 representation, $7 S_n \equiv 1 \pmod{2^n}$ is equivalent to $\left( x_{n-1} x_{n-2} \cdots x_1 x_0 000 \right)_2 - \left( x_{n-1} x_{n-2} \cdots x_1 x_0 \right)_2 - (1)_2 = \left( \cdots \underbrace{00\cdots 0}_{n \mbox{ digits} } \right)_2 .$

In the rest of the analysis, to lighten notation, we ease the base-2 subscription from all numbers. The equation above can be reformulated as:

\begin{table} \begin{tabular}{ccccccccc}

     &  & 0 &  & 0 & 0 & 0 & 0 & 0 \\
     &   &   &   &   &   &   &   & 1 \\
     &  &  &  &  &  &  &  &  \\
   \hline %or \bottomrule if using the `booktabs` package
     &    &  &  &  &  & 0 & 0 & 0\\
   \end{tabular}

\end{table}

Therefore, $x_0 = x_1 = x_2 = 1$ , $x_3 = 0$ , and for $k \geq 4$ , $x_k = x_{k-3}$ .

Therefore, $\begin{align*} x_{2019} + 2 x_{2020} + 4 x_{2021} + 8 x_{2022} & = x_3 + 2 x_1 + 4 x_2 + 8 x_3 \\ & = \boxed{\textbf{(A) 6}} . \end{align*}$

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

Solution 3

As in Solution $1$ , we note that the requested sum is $\frac{S_{2023}-S_{2019}}{2^{2019}}.$

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 requested 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

$\frac{S_{2023}-S_{2019}}{2^{2019}} = \frac{\frac{2^{2023}\cdot{x} + 1}{7} - \frac{2^{2019}\cdot{y} + 1}{7}}{2^{2019}}$

$= \frac{1}{2^{2019}}\left(\frac{2^{2023}\cdot{3} + 1}{7} -\left(\frac{2^{2019}\cdot{6} + 1}{7} \right)\right)$

$= \frac{3\cdot{2^4}-6}{7}$ $= \boxed{\textbf{(A) 6}}.$

-Benedict T (countmath1)

Video Solution

https://youtu.be/sBmk7tNSQBA

~ ThePuzzlr

https://youtu.be/2Dw75Zy6yAQ

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

Video Solution by OmegaLearn Using Binary and Modular Arithmetic

https://youtu.be/s_Bgj9srrXI

~ pi_is_3.14

2022 AMC 10B (Problems • Answer Key • Resources)
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2022 AMC 12B (Problems • Answer Key • Resources)
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