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

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(Video solution by grogg007 (No Binary))
 
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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  
 
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>
 
<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==
  
Suppose <math>7S_n \equiv 1 \pmod{2^n}</math> for all <math>n \geqslant 1</math>. What is the value of the sum
+
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>
<cmath>x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022}</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>\textbf{(A) } 6 \qquad \textbf{(B) } 7 \qquad \textbf{(C) }12\qquad \textbf{(D) } 14\qquad \textbf{(E) }15</math>
+
* <math>x_n=0</math> if and only if <math>n\equiv 0\pmod{3}.</math>
  
==Solution 1==
+
* <math>x_n=1</math> if and only if <math>n\not\equiv 0\pmod{3}.</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>
 
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 +
 +
~MRENTHUSIASM
  
 
==Solution 2==
 
==Solution 2==
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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>.
 
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>.
 
<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>
 
<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>
~ <math>\color{magenta} zoomanTV</math>
+
~zoomanTV
  
 
==Solution 3 ==
 
==Solution 3 ==
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-Benedict T (countmath1)
 
-Benedict T (countmath1)
  
==Video Solutions==
+
==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
 +
 
 +
==Solution 5 ==
 +
 
 +
Given that <math>7S_n \equiv 1 \pmod{2^n}</math>, we have
 +
<cmath>\begin{align*}
 +
7S_n &= a_n2^n + 1\\
 +
7S_{n-1} &= a_{n-1}2^{n-1} + 1.
 +
\end{align*}</cmath>
 +
where <math>a_n</math> is integer.
 +
 
 +
Notice that <cmath>S_n = S_{n-1} + 2^{n-1}x_{n-1}.</cmath>
 +
Thus,
 +
<cmath>\begin{align*}
 +
7(S_n - S_{n-1}) &= 2^{n-1}(2a_n - a_{n-1})\\
 +
7 \cdot 2^{n - 1}x_{n - 1} &= 2^{n - 1}(2a_n - a_{n - 1})\\
 +
7x_{n-1} &= 2a_n - a_{n-1}.
 +
\end{align*}</cmath>
 +
Obviously, <math>x_0 = 1</math>, <math>S_1 = 1</math>, <math>7S_1 = 3 \times 2^1 + 1</math>, so <math>a_1 = 3</math>.
 +
For each <math>i</math>, <math>x_i</math> must be 0 or 1, and <math>a_i</math> is an integer, so we can repeat the recursion to yield
 +
<cmath>\begin{align*}
 +
x_1 = \frac{2a_2 - a_1}{7} = \frac{2a_2 - 3}{7}\text{ can only be 1, where }a_2 = 5\\
 +
x_2 = \frac{2a_3 - a_2}{7} = \frac{2a_3 - 5}{7}\text{ can only be 1, where }a_3 = 6\\
 +
x_3 = \frac{2a_4 - a_3}{7} = \frac{2a_4 - 6}{7}\text{ can only be 0, where }a_4 = 3\\
 +
x_4 = \frac{2a_5 - a_4}{7} = \frac{2a_5 - 3}{7}\text{ can only be 1, where }a_5 = 5\\
 +
x_5 = \frac{2a_6 - a_5}{7} = \frac{2a_6 - 5}{7}\text{ can only be 1, where }a_6 = 6\\
 +
x_6 = \frac{2a_7 - a_6}{7} = \frac{2a_7 - 6}{7}\text{ can only be 0, where }a_7 = 3
 +
\end{align*}</cmath>
 +
So for any non-negative integer <math>k</math>,  we can find that <math>x_{3k + 1} = 1</math>, <math>x_{3k + 2} = 1</math>, <math>x_{3k + 3} = 0</math>, <cmath>x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022} = 0 + 2 \times 1 + 4 \times 1 + 8 \times  0 = \boxed{\textbf{(A) } 6}.</cmath>
 +
 
 +
~[https://artofproblemsolving.com/wiki/index.php/User:Reda_mandymath reda_mandymath]
 +
 
 +
==Solution 6==
 +
 
 +
Using the given conditions, it is evident that <math>x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022}=\frac{S_{2023}-S_{2019}}{2^{2019}}=a</math> (for simplicity). Moreover, because <math>7S_n \equiv 1 \pmod{2^n}</math> is true, an impulse to multiply the numerator and denominator by 7 is created. Let <math>7S_{2023}=2^{2023}k+1</math> and <math>7S_{2019}=2^{2019}k'+1</math>.
 +
\begin{align*}
 +
a=\frac{7S_{2023}-7S_{2019}}{7\cdot2^{2019}}&=\frac{2^{2023}k+1-2^{2019}k'-1}{7\cdot2^{2019}} \\
 +
&=\frac{2^4k-k'}{7} \\
 +
&=\frac{16k-k'}{7} \\
 +
\end{align*}
 +
Furthermore, because <math>0\le x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022}\le15</math>, <math>0\le\frac{16k-k'}{7}\le15</math>. The order pairs <math>(k,k',a)</math> could be found through trial and error that satisfies conditions <math>7S_{2023}=2^{2023}k+1</math> and <math>7S_{2019}=2^{2019}k'+1</math>. It is evident that <math>k\ne0,1,2</math> using Euler's Theorem.
 +
 
 +
<math>(3,6,6)</math> is valid because <math>7|2^{2023}\cdot3+1</math> and <math>7|2^{2019}\cdot6</math>.
 +
Thereby, <math>x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022}=\boxed{\textbf{(A) } 6}</math>.
 +
 
 +
~MaPhyCom
 +
 
 +
==Video Solution by MOP 2024==
 +
https://youtu.be/ShEE5WMhS2w
  
 +
~r00tsOfUnity
 +
 +
==Video Solution by ThePuzzlr==
 
https://youtu.be/sBmk7tNSQBA  
 
https://youtu.be/sBmk7tNSQBA  
  
 
~ ThePuzzlr
 
~ ThePuzzlr
  
 +
==Video Solution by Steven Chen==
 
https://youtu.be/2Dw75Zy6yAQ
 
https://youtu.be/2Dw75Zy6yAQ
  
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~ pi_is_3.14
 
~ pi_is_3.14
 +
 +
==Video Solution by The Power of Logic==
 +
https://youtu.be/rZaJSTbs7jY
 +
==Video Solution by Interstigation==
 +
https://youtu.be/r9VjnOzN4Ek
 +
 +
~Interstigation
 +
 +
==Video Solution by [[User:grogg007 | grogg007]] (No Binary)==
 +
https://youtu.be/EiJzaiw_YoQ?si=NvL9a8AslaH9m-IF
  
 
==See Also==
 
==See Also==

Latest revision as of 01:09, 5 July 2025

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

Solution 5

Given that $7S_n \equiv 1 \pmod{2^n}$, we have \begin{align*} 7S_n &= a_n2^n + 1\\ 7S_{n-1} &= a_{n-1}2^{n-1} + 1. \end{align*} where $a_n$ is integer.

Notice that \[S_n = S_{n-1} + 2^{n-1}x_{n-1}.\] Thus, \begin{align*} 7(S_n - S_{n-1}) &= 2^{n-1}(2a_n - a_{n-1})\\ 7 \cdot 2^{n - 1}x_{n - 1} &= 2^{n - 1}(2a_n - a_{n - 1})\\ 7x_{n-1} &= 2a_n - a_{n-1}. \end{align*} Obviously, $x_0 = 1$, $S_1 = 1$, $7S_1 = 3 \times 2^1 + 1$, so $a_1 = 3$. For each $i$, $x_i$ must be 0 or 1, and $a_i$ is an integer, so we can repeat the recursion to yield \begin{align*} x_1 = \frac{2a_2 - a_1}{7} = \frac{2a_2 - 3}{7}\text{ can only be 1, where }a_2 = 5\\ x_2 = \frac{2a_3 - a_2}{7} = \frac{2a_3 - 5}{7}\text{ can only be 1, where }a_3 = 6\\ x_3 = \frac{2a_4 - a_3}{7} = \frac{2a_4 - 6}{7}\text{ can only be 0, where }a_4 = 3\\ x_4 = \frac{2a_5 - a_4}{7} = \frac{2a_5 - 3}{7}\text{ can only be 1, where }a_5 = 5\\ x_5 = \frac{2a_6 - a_5}{7} = \frac{2a_6 - 5}{7}\text{ can only be 1, where }a_6 = 6\\ x_6 = \frac{2a_7 - a_6}{7} = \frac{2a_7 - 6}{7}\text{ can only be 0, where }a_7 = 3 \end{align*} So for any non-negative integer $k$, we can find that $x_{3k + 1} = 1$, $x_{3k + 2} = 1$, $x_{3k + 3} = 0$, \[x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022} = 0 + 2 \times 1 + 4 \times 1 + 8 \times  0 = \boxed{\textbf{(A) } 6}.\]

~reda_mandymath

Solution 6

Using the given conditions, it is evident that $x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022}=\frac{S_{2023}-S_{2019}}{2^{2019}}=a$ (for simplicity). Moreover, because $7S_n \equiv 1 \pmod{2^n}$ is true, an impulse to multiply the numerator and denominator by 7 is created. Let $7S_{2023}=2^{2023}k+1$ and $7S_{2019}=2^{2019}k'+1$. \begin{align*} a=\frac{7S_{2023}-7S_{2019}}{7\cdot2^{2019}}&=\frac{2^{2023}k+1-2^{2019}k'-1}{7\cdot2^{2019}} \\ &=\frac{2^4k-k'}{7} \\ &=\frac{16k-k'}{7} \\ \end{align*} Furthermore, because $0\le x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022}\le15$, $0\le\frac{16k-k'}{7}\le15$. The order pairs $(k,k',a)$ could be found through trial and error that satisfies conditions $7S_{2023}=2^{2023}k+1$ and $7S_{2019}=2^{2019}k'+1$. It is evident that $k\ne0,1,2$ using Euler's Theorem.

$(3,6,6)$ is valid because $7|2^{2023}\cdot3+1$ and $7|2^{2019}\cdot6$. Thereby, $x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022}=\boxed{\textbf{(A) } 6}$.

~MaPhyCom

Video Solution by MOP 2024

https://youtu.be/ShEE5WMhS2w

~r00tsOfUnity

Video Solution by ThePuzzlr

https://youtu.be/sBmk7tNSQBA

~ ThePuzzlr

Video Solution by Steven Chen

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

Video Solution by The Power of Logic

https://youtu.be/rZaJSTbs7jY

Video Solution by Interstigation

https://youtu.be/r9VjnOzN4Ek

~Interstigation

Video Solution by grogg007 (No Binary)

https://youtu.be/EiJzaiw_YoQ?si=NvL9a8AslaH9m-IF

See Also

2022 AMC 10B (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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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