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Difference between revisions of "2022 AMC 10B Problems/Problem 25"

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{{duplicate|[[2022 AMC 10B Problems#Problem 25|2022 AMC 10B #25]] and [[2022 AMC 12B Problems#Problem 23|2022 AMC 12B #23]]}}
 +
 
==Problem==
 
==Problem==
 +
Let <math>x_0,x_1,x_2,\dotsc</math> be a sequence of numbers, where each <math>x_k</math> is either <math>0</math> or <math>1</math>. For each positive integer <math>n</math>, define
 +
<cmath>S_n = \sum_{k=0}^{n-1} x_k 2^k</cmath>
 +
Suppose <math>7S_n \equiv 1 \pmod{2^n}</math> for all <math>n \geq 1</math>. What is the value of the sum 
 +
<cmath>x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022}?</cmath>
 +
<math>\textbf{(A) } 6 \qquad \textbf{(B) } 7 \qquad \textbf{(C) }12\qquad \textbf{(D) } 14\qquad \textbf{(E) }15</math>
 +
 +
==Solution 1==
 +
 +
In binary numbers, we have <cmath>S_n = (x_{n-1} x_{n-2} x_{n-3} x_{n-4} \ldots x_{2} x_{1} x_{0})_2.</cmath>
 +
It follows that <cmath>8S_n = (x_{n-1} x_{n-2} x_{n-3} x_{n-4} \ldots x_{2} x_{1} x_{0}000)_2.</cmath>
 +
We obtain <math>7S_n</math> by subtracting the equations:
 +
<cmath>\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_2 & x_1 & x_0)_2 \\
 +
\hline
 +
  & & & & & & & & & & &  \\ [-2.5ex]
 +
  & ( \ \ ?& ? & ? & 0 \ \ \ & \ldots & 0 & 0 & 0 & 0 & 0 & 1 \ )_2 \\
 +
\end{array}</cmath>
 +
We work from right to left:
 +
<cmath>\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*}</cmath>
 +
For all <math>n\geq3,</math> we conclude that
 +
 +
* <math>x_n=0</math> if and only if <math>n\equiv 0\pmod{3}.</math>
  
−
Let <math>x_0, x_1, x_2, \cdots</math> be a sequence of numbers, where each <math>x_k</math> is either 0 or 1. For each positive
+
* <math>x_n=1</math> if and only if <math>n\not\equiv 0\pmod{3}.</math>
−
integer <math>n</math>, define
 
−
<cmath>
 
−
\[
 
−
S_n = \sum_{k=0}^{n-1} x_k 2^k .
 
−
\]
 
−
</cmath>
 
  
−
Suppose <math>7 S_n \equiv 1 \pmod{2^n}</math> for all <math>n \geq 1</math>.
+
Finally, we get <math>(x_{2019},x_{2020},x_{2021},x_{2022})=(0,1,1,0),</math> from which <cmath>x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022} = \boxed{\textbf{(A) } 6}.</cmath>
−
What is the value of the sum
+
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
−
<cmath>
 
−
\[
 
−
x_{2019} + 2 x_{2020} + 4 x_{2021} + 8 x_{2022} ?
 
−
\]
 
−
</cmath>
 
  
−
==Solution (Base-2 Analysis)==
+
~MRENTHUSIASM
  
−
We solve this problem with base 2.
+
==Solution 2==
−
To avoid any confusion, for a base-2 number, we index the <math>k</math>th rightmost digit as digit <math>k-1</math>.
+
First, notice that <cmath>x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022} = \frac{S_{2023} - S_{2019}}{2^{2019}}.</cmath>
 +
Then since <math>S_n</math> is the modular inverse of <math>7</math> in <math>\mathbb{Z}_{2^n}</math>, we can perform the Euclidean Algorithm to find it for <math>n = 2019,2023</math>.
  
−
We have <math>S_n = \left( x_{n-1} x_{n-2} \cdots x_1 x_0 \right)_2</math>.
+
Starting with <math>2019</math>,
 +
<cmath>\begin{align*}
 +
7S_{2019} &\equiv 1 \pmod{2^{2019}} \\
 +
7S_{2019} &= 2^{2019}k + 1.
 +
\end{align*}</cmath>
 +
Now, take both sides <math>\operatorname{mod} \ 7</math>:
 +
<cmath>0 \equiv 2^{2019}k + 1 \pmod{7}.</cmath>
 +
Using Fermat's Little Theorem, <cmath>2^{2019} = (2^{336})^6 \cdot 2^3 \equiv 2^3 \equiv 1 \pmod{7}.</cmath>
 +
Thus,  <cmath>0 \equiv k + 1 \pmod{7} \implies k \equiv 6 \pmod{7} \implies k = 7j + 6.</cmath>
 +
Therefore, <cmath>7S_{2019} = 2^{2019} (7j + 6) + 1 \implies S_{2019} = \frac{2^{2019} (7j + 6) + 1}{7}.</cmath>
  
−
In the base-2 representation, <math>7 S_n \equiv 1 \pmod{2^n}</math> is equivalent to
+
We may repeat this same calculation with <math>S_{2023}</math> to yield <cmath>S_{2023} = \frac{2^{2023} (7h + 3) + 1}{7}.</cmath>
−
<cmath>
+
Now, we notice that <math>S_n</math> is basically an integer expressed in binary form with <math>n</math> bits.
−
\[
+
This gives rise to a simple inequality, <cmath>0 \leqslant S_n \leqslant 2^n.</cmath>
−
\left( x_{n-1} x_{n-2} \cdots x_1 x_0 000 \right)_2
+
Since the maximum possible number that can be generated with <math>n</math> bits is <cmath>\underbrace{{11111\dotsc1}_2}_{n} = \sum_{k=0}^{n-1} 2^k = 2^n - 1 \leqslant 2^n.</cmath>
−
- \left( x_{n-1} x_{n-2} \cdots x_1 x_0 \right)_2
+
Looking at our calculations for <math>S_{2019}</math> and <math>S_{2023}</math>, we see that the only valid integers that satisfy that constraint are <math>j = h = 0</math>.
−
- (1)_2
+
<cmath>\frac{S_{2023} - S_{2019}}{2^{2019}} = \frac{\tfrac{2^{2023} \cdot 3 + 1}{7} - \tfrac{2^{2019} \cdot 6 + 1}{7}}{2^{2019}} = \frac{2^4 \cdot 3 - 6}{7} = \boxed{\textbf{(A) } 6}.</cmath>
−
= \left( \cdots \underbrace{00\cdots 0}_{n \mbox{ digits} } \right)_2 .
+
~zoomanTV
−
\]
 
−
</cmath>
 
  
−
In the rest of the analysis, to lighten notation, we ease the base-2 subscription from all numbers.
+
==Solution 3 ==
−
The equation above can be reformulated as:
 
  
−
<math></math>
+
As in Solution 2, we note that
−
\begin{tabular}{ccccccccc}
+
<cmath>x_{2019}+2x_{2020}+4x_{2021}+8x_{2022}=\frac{S_{2023}-S_{2019}}{2^{2019}}.</cmath>
−
      & <math>\cdots</math> & 0 & <math>\cdots</math> & 0 & 0 & 0 & 0 & 0 \\
+
We also know that <math>7S_{2023} \equiv 1 \pmod{2^{2023}}</math> and <math>7S_{2019} \equiv 1 \pmod{2^{2019}}</math>, this implies:
−
      &  &  &  &  &  &  &  & 1 \\
+
<cmath> \textbf{(1) } 7S_{2023}=2^{2023}\cdot{x} + 1, </cmath>
−
      <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> \\
+
<cmath> \textbf{(2) } 7S_{2019}=2^{2019}\cdot{y} + 1. </cmath>
−
    \hline %or \bottomrule if using the `booktabs` package
+
Dividing by <math>7</math>, we can isolate the previous sums:
−
      & <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\\
+
<cmath> \textbf{(3) } S_{2023}=\frac{2^{2023}\cdot{x} + 1}{7}, </cmath>
−
    \end{tabular}
+
<cmath> \textbf{(4) } S_{2019}=\frac{2^{2019}\cdot{y} + 1}{7}. </cmath>
−
<math></math>
+
The maximum value of <math>S_n</math> occurs when every <math>x_i</math> is equal to <math>1</math>. Even when this happens, the value of <math>S_n</math> is less than <math>2^n</math>. Therefore, we can construct the following inequalities:
 +
<cmath> \textbf{(3) } S_{2023}=\frac{2^{2023}\cdot{x} + 1}{7} < 2^{2023}, </cmath>
 +
<cmath> \textbf{(4) } S_{2019}=\frac{2^{2019}\cdot{y} + 1}{7} < 2^{2019}. </cmath>
 +
From these two equations, we can deduce that both <math>x</math> and <math>y</math> are less than <math>7</math>.
  
−
Therefore, <math>x_0 = x_1 = x_2 = 1</math>, <math>x_3 = 0</math>, and for <math>k \geq 4</math>, <math>x_k = x_{k-3}</math>.
+
Reducing <math>\textbf{1}</math> and <math>\textbf{2}</math> <math>\pmod{7},</math> we see that
 +
<cmath>2^{2023}\cdot{x}\equiv 6\pmod{7},</cmath>
 +
and
 +
<cmath>2^{2019}\cdot{y}\equiv 6\pmod{7}.</cmath>
  
−
Therefore,
+
The powers of <math>2</math> repeat every <math>3, \pmod{7}.</math>  
−
<cmath>
 
−
\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*}
 
−
</cmath>
 
  
−
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
+
Therefore, <math>2^{2023}\equiv 2 \pmod 7</math> and <math>2^{2019} \equiv 1 \pmod {7}.</math> Substituing this back into the above equations,
 +
<cmath>2x\equiv{6}\pmod{7}</cmath>
 +
and
 +
<cmath>y\equiv{6}\pmod{7}.</cmath>
 +
 
 +
Since <math>x</math> and <math>y</math> are integers less than <math>7</math>, the only values of <math>x</math> and <math>y</math> are <math>3</math> and <math>6</math> respectively.
 +
 
 +
The requested sum is
 +
<cmath>\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*}</cmath>
 +
-Benedict T (countmath1)
 +
 
 +
==Solution 4 ==
 +
 
 +
Note that, as in Solution 2, we have <cmath>x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022} = \frac{S_{2023} - S_{2019}}{2^{2019}}.</cmath> This is because <cmath>S_{2023} = x_{0}2^{0} + x_{1}2^{1} + \cdots + x_{2019}2^{2019} + \cdots + x_{2022}2^{2022}</cmath> and <cmath>S_{2019} = x_{0}2^{0} + x_{1}2^{1} + \cdots + x_{2018}2^{2018}.</cmath> Note that <cmath>S_{2023} - S_{2019} = x_{2019}2^{2019} + \cdots + x_{2022}2^{2022} = 2^{2019}(x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022}).</cmath> Therefore, <cmath>x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022} = \frac{S_{2023} - S_{2019}}{2^{2019}}.</cmath> Multiplying both sides by 7 gives us <cmath>7(x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022}) = \frac{7S_{2023} - 7S_{2019}}{2^{2019}}.</cmath> We can write <cmath>7S_{2023} = 1\pmod{2^{2023}} = 1 + 2^{2023}a = 1 + 2^{2019}*16a</cmath> and <cmath>7S_{2019} = 1\pmod{2^{2019}} = 1 + 2^{2019}b</cmath> for some a and b. Substituting, we get <cmath>7(x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022}) = \frac{(1 + 2^{2019} * 16a) - (1 + 2^{2019}b)}{2^{2019}} = 16a - b.</cmath> Therefore, our answer can be written as <cmath>\frac{16a - b}{7}.</cmath> Another thing to notice is that a and b are integers between 0 and 6. This is because <cmath>7(1 + 2 + 4 + 8 + \cdots + 2^{2022}) \geqslant 7S_{2023} = 1 + 2^{2023}a</cmath> which is <cmath>7(2^{2023}) - 7 \geqslant 1 + 2^{2023}a</cmath> <cmath>(7-a) \geqslant \frac{8}{2^{2023}}</cmath> which only holds when a is less than 7 because the right is very small positive number, so the left must be positive, too. Clearly, a is also non-negative, because otherwise, <cmath>7S_{2023} = 1 + 2^{2023}a < 0</cmath> which would mean <cmath>S_{2023} < 0</cmath> which cannot happen, so a is greater than 0.  A similar explanation for b shows that b is an integer between 0 and 6 inclusive.
 +
 
 +
Going back to the solution, if our answer to the problem is n, then <cmath>16a - b = 7n</cmath> and <cmath>16a = 7n + b,</cmath> so we can try the five option choices and see which one, when multiplied by 7 and added to some whole number between 0 and 6 results in a multiple of 16. Trying all the option choices, we see that you need to add 7n to something more than 6 to equal a multiple of 16 other than for option A. Therefore, the answer is <math>\boxed{\textbf{(A) } 6}.</math>
 +
 
 +
-Rutvik Arora (youtube channel: https://www.youtube.com/channel/UCkgAgmNAQV8WGTOazGYEGwg)
 +
-whatdohumanitarianseat made a small edit for a typo
 +
 
 +
==Video Solution by MOP 2024==
 +
https://youtu.be/ShEE5WMhS2w
 +
 
 +
~r00tsOfUnity
 +
 
 +
==Video Solution by ThePuzzlr==
 +
https://youtu.be/sBmk7tNSQBA
  
−
==Video Solution==
+
~ ThePuzzlr
  
 +
==Video Solution by Steven Chen==
 
https://youtu.be/2Dw75Zy6yAQ
 
https://youtu.be/2Dw75Zy6yAQ
  
 
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 
~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
 +
 +
==Video Solution by The Power of Logic==
 +
https://youtu.be/rZaJSTbs7jY
 +
==Video Solution by Interstigation==
 +
https://youtu.be/r9VjnOzN4Ek
 +
 +
~Interstigation
 +
 +
==See Also==
 +
{{AMC10 box|year=2022|ab=B|num-b=24|after=Last problem}}
 +
{{AMC12 box|year=2022|ab=B|num-b=22|num-a=24}}
 +
 +
[[Category:Intermediate Number Theory Problems]]
 +
{{MAA Notice}}

Latest revision as of 02:13, 30 December 2023

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

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: \[\begin{array}{clccrccccccr} & (x_{n-1} & x_{n-2} & x_{n-3} & x_{n-4} & \ldots & x_2 & x_1 & x_0 & 0 & 0 & 0 \ )_2 \\ -\quad & & & & (x_{n-1} & \ldots & x_5 & x_4 & x_3 & x_2 & x_1 & x_0)_2 \\ \hline & & & & & & & & & & & \\ [-2.5ex] & ( \ \ ?& ? & ? & 0 \ \ \ & \ldots & 0 & 0 & 0 & 0 & 0 & 1 \ )_2 \\ \end{array}\] 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 $j = h = 0$. \[\frac{S_{2023} - S_{2019}}{2^{2019}} = \frac{\tfrac{2^{2023} \cdot 3 + 1}{7} - \tfrac{2^{2019} \cdot 6 + 1}{7}}{2^{2019}} = \frac{2^4 \cdot 3 - 6}{7} = \boxed{\textbf{(A) } 6}.\] ~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)

Solution 4

Note that, as in Solution 2, we have \[x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022} = \frac{S_{2023} - S_{2019}}{2^{2019}}.\] This is because \[S_{2023} = x_{0}2^{0} + x_{1}2^{1} + \cdots + x_{2019}2^{2019} + \cdots + x_{2022}2^{2022}\] and \[S_{2019} = x_{0}2^{0} + x_{1}2^{1} + \cdots + x_{2018}2^{2018}.\] Note that \[S_{2023} - S_{2019} = x_{2019}2^{2019} + \cdots + x_{2022}2^{2022} = 2^{2019}(x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022}).\] Therefore, \[x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022} = \frac{S_{2023} - S_{2019}}{2^{2019}}.\] Multiplying both sides by 7 gives us \[7(x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022}) = \frac{7S_{2023} - 7S_{2019}}{2^{2019}}.\] We can write \[7S_{2023} = 1\pmod{2^{2023}} = 1 + 2^{2023}a = 1 + 2^{2019}*16a\] and \[7S_{2019} = 1\pmod{2^{2019}} = 1 + 2^{2019}b\] for some a and b. Substituting, we get \[7(x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022}) = \frac{(1 + 2^{2019} * 16a) - (1 + 2^{2019}b)}{2^{2019}} = 16a - b.\] Therefore, our answer can be written as \[\frac{16a - b}{7}.\] Another thing to notice is that a and b are integers between 0 and 6. This is because \[7(1 + 2 + 4 + 8 + \cdots + 2^{2022}) \geqslant 7S_{2023} = 1 + 2^{2023}a\] which is \[7(2^{2023}) - 7 \geqslant 1 + 2^{2023}a\] \[(7-a) \geqslant \frac{8}{2^{2023}}\] which only holds when a is less than 7 because the right is very small positive number, so the left must be positive, too. Clearly, a is also non-negative, because otherwise, \[7S_{2023} = 1 + 2^{2023}a < 0\] which would mean \[S_{2023} < 0\] which cannot happen, so a is greater than 0. A similar explanation for b shows that b is an integer between 0 and 6 inclusive.

Going back to the solution, if our answer to the problem is n, then \[16a - b = 7n\] and \[16a = 7n + b,\] so we can try the five option choices and see which one, when multiplied by 7 and added to some whole number between 0 and 6 results in a multiple of 16. Trying all the option choices, we see that you need to add 7n to something more than 6 to equal a multiple of 16 other than for option A. Therefore, the answer is $\boxed{\textbf{(A) } 6}.$

-Rutvik Arora (youtube channel: https://www.youtube.com/channel/UCkgAgmNAQV8WGTOazGYEGwg) -whatdohumanitarianseat made a small edit for a typo

Video Solution by MOP 2024

https://youtu.be/ShEE5WMhS2w

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Video Solution by ThePuzzlr

https://youtu.be/sBmk7tNSQBA

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Video Solution by Steven Chen

https://youtu.be/2Dw75Zy6yAQ

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Video Solution by OmegaLearn Using Binary and Modular Arithmetic

https://youtu.be/s_Bgj9srrXI

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Video Solution by The Power of Logic

https://youtu.be/rZaJSTbs7jY

Video Solution by Interstigation

https://youtu.be/r9VjnOzN4Ek

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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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