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

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#REDIRECT [[2022_AMC_12B_Problems/Problem_23]]
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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==
 +
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 \geqslant 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==
 +
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>.
 +
 
 +
Starting with <math>2019</math>, <cmath>7S_{2019} \equiv 1 \pmod{2^{2019}}</cmath>
 +
<cmath>7S_{2019} = 2^{2019}k + 1</cmath>
 +
Now, take both sides <math>\text{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>
 +
 
 +
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>
 +
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>
 +
<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
 +
 
 +
<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{(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{(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{(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>
 +
and
 +
<cmath>2^{2019}\cdot{y}\equiv 6\pmod{7}.</cmath>
 +
 
 +
The powers of <math>2</math> repeat every <math>3, \pmod{7}.</math>
 +
 
 +
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>\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>
 +
 
 +
<cmath> = \frac{1}{2^{2019}}\left(\frac{2^{2023}\cdot{3} + 1}{7} -\left(\frac{2^{2019}\cdot{6} + 1}{7}  \right)\right) </cmath>
 +
 
 +
<cmath> = \frac{3\cdot{2^4}-6}{7}</cmath>
 +
<cmath> = \boxed{\textbf{(A) 6}}.</cmath>
 +
 
 +
-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
 +
 
 +
==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}}

Revision as of 07:58, 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

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 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

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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