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Difference between revisions of "2022 AMC 12B Problems/Problem 23"

m
Line 31: Line 31:
 
<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>
 
~ <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 %or \bottomrule if using the `booktabs` package
 +
      & <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)
 +
 +
==Video Solution==
 +
 +
https://youtu.be/2Dw75Zy6yAQ
 +
 +
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
  
 
==See Also==
 
==See Also==

Revision as of 16:50, 17 November 2022

The following problem is from both the 2022 AMC 12B #23 and 2022 AMC 10B #25, 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 \[x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022}\]


$\textbf{(A)}~6\qquad\textbf{(B)}~7\qquad\textbf{(C)}~12\qquad\textbf{(D)}~14\qquad\textbf{(E)}~15\qquad$

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 $\mathbb{Z}_{n}$, 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, \[2^{2019} = (2^{288})^7 \cdot 2^3 \equiv 2^3 \equiv 1 \pmod{7}\] 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 %or \bottomrule if using the `booktabs` package
     & $x_{n-1}$ $x_{n-2}$ $x_{n-3}$ & $x_{n-4}$ & $\cdots$ & $x_1$ & $x_0$ & 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)

Video Solution

https://youtu.be/2Dw75Zy6yAQ

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

See Also

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
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

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