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

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{{duplicate|[[2022 AMC 12B Problems#Problem 23|2022 AMC 12B #23]] and [[2022 AMC 10B Problems#Problem 25|2022 AMC 10B #25]]}}
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#REDIRECT [[2022_AMC_10B_Problems/Problem_25]]
 
 
==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\qquad</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}_{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^{288})^7 \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 %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==
 
{{AMC12 box|year=2022|ab=B|num-b=22|num-a=24}}
 
{{AMC10 box|year=2022|ab=B|num-b=24|after=Last problem}}
 
{{MAA Notice}}
 

Latest revision as of 06:57, 5 December 2022