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

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7S_{n-1} &= a_{n-1}2^{n-1} + 1.
 
7S_{n-1} &= a_{n-1}2^{n-1} + 1.
 
\end{align*}</cmath>
 
\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>
 
Notice that <cmath>S_n = S_{n-1} + 2^{n-1}x_{n-1}.</cmath>
 
Thus,
 
Thus,
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
 
7(S_n - S_{n-1}) &= 2^{n-1}(2a_n - a_{n-1})\\
 
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}.
 
7x_{n-1} &= 2a_n - a_{n-1}.
−
\end{align*}</cmath>
 
−
Therefore,
 
−
<cmath>\begin{align*}
 
−
x_{n-1} &\equiv a_{n-1} \pmod{2}\\
 
−
a_n &= \frac{7x_{n-1} + a_{n-1}}{2}.
 
 
\end{align*}</cmath>
 
\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>.  
 
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>.  
−
We repeat the recursion to yield
+
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*}
 
<cmath>\begin{align*}
−
x_1 = 1 \equiv a_1 \pmod{2}, \qquad a_2 = \frac{7x_{1} + a_{1}}{2} = 5\\
+
x_1 = \frac{2a_2 - a_1}{7} = \frac{2a_2 - 3}{7}\text{ can only be 1, where }a_2 = 5\\
−
x_2 = 1 \equiv a_2 \pmod{2}, \qquad a_3 = \frac{7x_{1} + a_{1}}{2} = 6\\
+
x_2 = \frac{2a_3 - a_2}{7} = \frac{2a_3 - 5}{7}\text{ can only be 1, where }a_3 = 6\\
−
x_3 = 0 \equiv a_3 \pmod{2}, \qquad a_4 = \frac{7x_{1} + a_{1}}{2} = 3\\
+
x_3 = \frac{2a_4 - a_3}{7} = \frac{2a_4 - 6}{7}\text{ can only be 0, where }a_4 = 3\\
−
x_4 = 1 \equiv a_4 \pmod{2}, \qquad a_5 = \frac{7x_{1} + a_{1}}{2} = 5\\
+
x_4 = \frac{2a_5 - a_4}{7} = \frac{2a_5 - 3}{7}\text{ can only be 1, where }a_5 = 5\\
−
x_5 = 1 \equiv a_5 \pmod{2}, \qquad a_6 = \frac{7x_{1} + a_{1}}{2} = 6\\
+
x_5 = \frac{2a_6 - a_5}{7} = \frac{2a_6 - 5}{7}\text{ can only be 1, where }a_6 = 6\\
−
x_6 = 0 \equiv a_6 \pmod{2}, \qquad a_7 = \frac{7x_{1} + a_{1}}{2} = 3
+
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>
 
\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>
 
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>
  
−
~reda
+
~[https://artofproblemsolving.com/wiki/index.php/User:Reda_mandymath reda_mandymath]
  
 
==Video Solution by MOP 2024==
 
==Video Solution by MOP 2024==

Latest revision as of 10:25, 8 October 2024

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

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

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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