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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]]}}
| + | #REDIRECT [[2022_AMC_10B_Problems/Problem_25]] |
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| − | ==Problem==
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| − | 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
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| − | <cmath>S_n = \sum_{k=0}^{n-1} x_k 2^k</cmath>
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| − | Suppose <math>7S_n \equiv 1 \pmod{2^n}</math> for all <math>n \geqslant 1</math>. What is the value of the sum
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| − | <cmath>x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022}</cmath>
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| − | <math>\textbf{(A)}~6\qquad\textbf{(B)}~7\qquad\textbf{(C)}~12\qquad\textbf{(D)}~14\qquad\textbf{(E)}~15\qquad</math>
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| − | ==Solution==
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| − | First, notice that <cmath>x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022} = \frac{S_{2023} - S_{2019}}{2^{2019}}</cmath>
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| − | 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>.
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| − | Starting with <math>2019</math>, <cmath>7S_{2019} \equiv 1 \pmod{2^{2019}}</cmath>
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| − | <cmath>7S_{2019} = 2^{2019}k + 1</cmath>
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| − | Now, take both sides <math>\text{mod } 7</math>
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| − | <cmath>0 \equiv 2^{2019}k + 1 \pmod{7}</cmath>
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| − | Using Fermat's Little Theorem, <cmath>2^{2019} = (2^{336})^6 \cdot 2^3 \equiv 2^3 \equiv 1 \pmod{7}</cmath>
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| − | Thus, <cmath>0 \equiv k + 1 \pmod{7} \implies k \equiv 6 \pmod{7} \implies k = 7j + 6</cmath>
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| − | Therefore, <cmath>7S_{2019} = 2^{2019} (7j + 6) + 1 \implies S_{2019} = \frac{2^{2019} (7j + 6) + 1}{7}</cmath>
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| − | We may repeat this same calculation with <math>S_{2023}</math> to yield <cmath>S_{2023} = \frac{2^{2023} (7h + 3) + 1}{7}</cmath>
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| − | Now, we notice that <math>S_n</math> is basically an integer expressed in binary form with <math>n</math> bits.
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| − | This gives rise to a simple inequality, <cmath>0 \leqslant S_n \leqslant 2^n</cmath>
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| − | 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>
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| − | 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>
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| − | <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>
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| − | ~ <math>\color{magenta} zoomanTV</math>
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| − | ==Solution 2 (Base-2 Analysis)==
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| − | We solve this problem with base 2.
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| − | To avoid any confusion, for a base-2 number, we index the <math>k</math>th rightmost digit as digit <math>k-1</math>.
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| − | We have <math>S_n = \left( x_{n-1} x_{n-2} \cdots x_1 x_0 \right)_2</math>.
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| − | In the base-2 representation, <math>7 S_n \equiv 1 \pmod{2^n}</math> is equivalent to
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| − | <cmath>
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| − | \[
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| − | \left( x_{n-1} x_{n-2} \cdots x_1 x_0 000 \right)_2
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| − | - \left( x_{n-1} x_{n-2} \cdots x_1 x_0 \right)_2
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| − | - (1)_2
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| − | = \left( \cdots \underbrace{00\cdots 0}_{n \mbox{ digits} } \right)_2 .
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| − | \]
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| − | </cmath>
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| − | In the rest of the analysis, to lighten notation, we ease the base-2 subscription from all numbers.
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| − | The equation above can be reformulated as:
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| − | \begin{table}
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| − | \begin{tabular}{ccccccccc}
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| − | & <math>\cdots</math> & 0 & <math>\cdots</math> & 0 & 0 & 0 & 0 & 0 \\
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| − | & & & & & & & & 1 \\
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| − | <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> \\
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| − | \hline %or \bottomrule if using the `booktabs` package
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| − | & <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\\
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| − | \end{tabular}
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| − | \end{table}
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| − | 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>.
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| − | Therefore,
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| − | <cmath>
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| − | \begin{align*}
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| − | x_{2019} + 2 x_{2020} + 4 x_{2021} + 8 x_{2022}
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| − | & = x_3 + 2 x_1 + 4 x_2 + 8 x_3 \\
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| − | & = \boxed{\textbf{(A) 6}} .
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| − | \end{align*}
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| − | </cmath>
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| − | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
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| − | ==Video Solution==
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| − | https://youtu.be/sBmk7tNSQBA
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| − | ~ ThePuzzlr
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| − | https://youtu.be/2Dw75Zy6yAQ
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| − | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
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| − | ==See Also==
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| − | {{AMC12 box|year=2022|ab=B|num-b=22|num-a=24}}
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| − | {{AMC10 box|year=2022|ab=B|num-b=24|after=Last problem}}
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| − | [[Category:Intermediate Number Theory Problems]]
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| − | {{MAA Notice}}
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