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Difference between revisions of "2022 AMC 12B Problems/Problem 23"

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       <math>+</math>&  & <math>x_{n-1}</math> & <math>\cdots</math> & <math>x_4</math> & <math>x_3</math> & <math>x_2</math> & <math>x_1</math> & <math>x_0</math> \\
 
       <math>+</math>&  & <math>x_{n-1}</math> & <math>\cdots</math> & <math>x_4</math> & <math>x_3</math> & <math>x_2</math> & <math>x_1</math> & <math>x_0</math> \\
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       & <math>x_{n-1}</math> <math>x_{n-2}</math> <math>x_{n-3}</math> & <math>x_{n-4}</math> & <math>\cdots</math> & <math>x_1</math> & <math>x_0</math> & 0 & 0 & 0\\
 
       & <math>x_{n-1}</math> <math>x_{n-2}</math> <math>x_{n-3}</math> & <math>x_{n-4}</math> & <math>\cdots</math> & <math>x_1</math> & <math>x_0</math> & 0 & 0 & 0\\
 
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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
−
 
  
 
==Solution 3 ==
 
==Solution 3 ==

Revision as of 13:00, 20 November 2022

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} 

     & $\cdots$ & 0 & $\cdots$ & 0 & 0 & 0 & 0 & 0 \\
     &   &   &   &   &   &   &   & 1 \\
     $+$&  & $x_{n-1}$ & $\cdots$ & $x_4$ & $x_3$ & $x_2$ & $x_1$ & $x_0$ \\
   \hline
     & $x_{n-1}$ $x_{n-2}$ $x_{n-3}$ & $x_{n-4}$ & $\cdots$ & $x_1$ & $x_0$ & 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

\[x_{2019}+2x_{2020}+4x_{2021}+8x_{2022}=\frac{S_{2023}-S_{2019}}{2^{2019}}.\]

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

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

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

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

-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

See Also

2022 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Last problem
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions
2022 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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