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

Revision as of 08:48, 5 December 2022

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 $n \geqslant 1$ . 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

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

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 $j = h = 0$ . $\frac{S_{2023} - S_{2019}}{2^{2019}} = \frac{\tfrac{2^{2023} \cdot 3 + 1}{7} - \tfrac{2^{2019} \cdot 6 + 1}{7}}{2^{2019}} = \frac{2^4 \cdot 3 - 6}{7} = \boxed{\textbf{(A) } 6}.$ ~ $\color{magenta} zoomanTV$

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)

Video Solutions

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

@@ Line 10: / Line 10: @@
 <math>\textbf{(A) } 6 \qquad \textbf{(B) } 7 \qquad \textbf{(C) }12\qquad \textbf{(D) } 14\qquad \textbf{(E) }15</math>
-==Solution==
+==Solution 1==
-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 7 in <math>\mathbb{Z}_{2^n}</math> , we can perform the Euclidean algorithm to find it for <math>n = 2019,2023</math>.
+~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
+==Solution 2==
+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>.
-Starting with <math>2019</math>, <cmath>7S_{2019} \equiv 1 \pmod{2^{2019}}</cmath>
+Starting with <math>2019</math>,
-<cmath>7S_{2019} = 2^{2019}k + 1</cmath>
+<cmath>\begin{align*}
-Now, take both sides <math>\text{mod } 7</math>
+S_{2019} &\equiv 1 \pmod{2^{2019}} \\
-<cmath>0 \equiv 2^{2019}k + 1 \pmod{7}</cmath>
+S_{2019} &= 2^{2019}k + 1.
-Using Fermat's Little Theorem, <cmath>2^{2019} = (2^{336})^6 \cdot 2^3 \equiv 2^3 \equiv 1 \pmod{7}</cmath>
+\end{align*}</cmath>
-Thus,  <cmath>0 \equiv k + 1 \pmod{7} \implies k \equiv 6 \pmod{7} \implies k = 7j + 6</cmath>
+Now, take both sides <math>\operatorname{mod} \ 7</math>:
-Therefore, <cmath>7S_{2019} = 2^{2019} (7j + 6) + 1 \implies S_{2019} = \frac{2^{2019} (7j + 6) + 1}{7}</cmath>
+<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>
-We may repeat this same calculation with <math>S_{2023}</math> to yield <cmath>S_{2023} = \frac{2^{2023} (7h + 3) + 1}{7}</cmath>
+We may repeat this same calculation with <math>S_{2023}</math> to yield <cmath>S_{2023} = \frac{2^{2023} (7h + 3) + 1}{7}.</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>
+This gives rise to a simple inequality, <cmath>0 \leqslant S_n \leqslant 2^n.</cmath>
-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>
+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>
-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>
-==Solution 2 (Base-2 Analysis)==
-We solve this problem with base 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>.
-We have <math>S_n = \left( x_{n-1} x_{n-2} \cdots x_1 x_0 \right)_2</math>.
-In the base-2 representation, <math>7 S_n \equiv 1 \pmod{2^n}</math> is equivalent to
-<cmath>
-\[
-\left( x_{n-1} x_{n-2} \cdots x_1 x_0 000 \right)_2
-- \left( x_{n-1} x_{n-2} \cdots x_1 x_0 \right)_2
-- (1)_2
-= \left( \cdots \underbrace{00\cdots 0}_{n \mbox{ digits} } \right)_2 .
-\]
-</cmath>
-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}
-      & <math>\cdots</math> & 0 & <math>\cdots</math> & 0 & 0 & 0 & 0 & 0 \\
-      &   &   &   &   &   &   &   & 1 \\
-      <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> \\
-    \hline
-      & <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\\
-\end{tabular}
-\end{table}
-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>.
-Therefore,
-<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)
 ==Solution 3 ==
-As in Solution <math>1</math>, we note that
+As in Solution 2, we note that
 <cmath>x_{2019}+2x_{2020}+4x_{2021}+8x_{2022}=\frac{S_{2023}-S_{2019}}{2^{2019}}.</cmath>
 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:
+<cmath> \textbf{(1) } 7S_{2023}=2^{2023}\cdot{x} + 1, </cmath>
-<cmath> \textbf{(1) } 7S_{2023}=2^{2023}\cdot{x} + 1 </cmath>
 <cmath> \textbf{(2) } 7S_{2019}=2^{2019}\cdot{y} + 1. </cmath>
 Dividing by <math>7</math>, we can isolate the previous sums:
+<cmath> \textbf{(3) } S_{2023}=\frac{2^{2023}\cdot{x} + 1}{7}, </cmath>
-<cmath> \textbf{(3) } S_{2023}=\frac{2^{2023}\cdot{x} + 1}{7} </cmath>
 <cmath> \textbf{(4) } S_{2019}=\frac{2^{2019}\cdot{y} + 1}{7}. </cmath>
 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{(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>.
 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>
-<cmath>2^{2023}\cdot{x}\equiv 6\pmod{7}</cmath>
 and
 <cmath>2^{2019}\cdot{y}\equiv 6\pmod{7}.</cmath>
@@ Line 106: / Line 60: @@
 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
@@ Line 114: / Line 67: @@
 The requested sum is
+<cmath>\begin{align*}
-<cmath>\frac{S_{2023}-S_{2019}}{2^{2019}} = \frac{\frac{2^{2023}\cdot{x} + 1}{7} - \frac{2^{2019}\cdot{y} + 1}{7}}{2^{2019}}</cmath>
+\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) \\
-<cmath> = \frac{1}{2^{2019}}\left(\frac{2^{2023}\cdot{3} + 1}{7} -\left(\frac{2^{2019}\cdot{6} + 1}{7}  \right)\right) </cmath>
+&= \frac{3\cdot{2^4}-6}{7} \\
+&= \boxed{\textbf{(A) } 6}.
-<cmath> = \frac{3\cdot{2^4}-6}{7}</cmath>
+\end{align*}</cmath>
-<cmath> = \boxed{\textbf{(A) 6}}.</cmath>
 -Benedict T (countmath1)

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