67 lines
2.1 KiB
TeX
67 lines
2.1 KiB
TeX
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\documentclass[12pt,a4paper,german]{article}
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\usepackage{url}
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%\usepackage{graphics}
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\usepackage{times}
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\usepackage[T1]{fontenc}
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\usepackage{ngerman}
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\usepackage{float}
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\usepackage{diagbox}
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\usepackage[utf8]{inputenc}
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\usepackage{geometry}
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\usepackage{amsfonts}
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\usepackage{amsmath}
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\usepackage{cancel}
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\usepackage{wasysym}
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\usepackage{csquotes}
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\usepackage{graphicx}
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\usepackage{epsfig}
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\usepackage{paralist}
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\usepackage{tikz}
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\geometry{left=2.0cm,textwidth=17cm,top=3.5cm,textheight=23cm}
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%%%%%%%%%% Fill out the the definitions %%%%%%%%%
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\def \name {Valentin Brandl} %
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\def \matrikel {108018274494} %
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\def \pname {Marvin Herrmann} %
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\def \pmatrikel {108018265436} %
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\def \gruppe {2 (Mi 10-12 Andre)}
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\def \qname {Pascal Brackmann}
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\def \qmatrikel {108017113834} %
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\def \uebung {10} %
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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% DO NOT MODIFY THIS HEADER
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\newcommand{\hwsol}{
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\vspace*{-2cm}
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\noindent \matrikel \quad \name \hfill \"Ubungsgruppe: \gruppe \\
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\noindent \pmatrikel \quad \pname \\
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\noindent \qmatrikel \quad \qname \\
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\begin{center}{\Large \bf L\"osung f\"ur \"Ubung \# \uebung}\end{center}
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}
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\begin{document}
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%Import header
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\hwsol
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\section*{Aufgabe 10.1}
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Kriterium:
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Eine Zahl $z \in \mathbb{N}$ ist durch 37 teilbar, wenn ihre nicht-alternierende hunderter-Quersumme durch 37 teilbar ist.\\
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\\
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Beweis: Sei $z=a_0+a_1 \cdot 10 + a_2 \cdot 10^2 + ... + a_n \cdot 10^n$.\\
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Es gilt $37 |z \Leftrightarrow z \equiv 0 \text{ mod 37 }$\\
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$\Rightarrow a_0+a_1\cdot 10 + a_2 \cdot 10^2 + ... + a_n \cdot 10^n \equiv 0 \text{ mod 37}$\\
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Es gilt nun $10^3 \equiv 1 \text{ mod } 37$\\
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\\
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$\Rightarrow a_0 + a_1 \cdot 10^1 + a_2 \cdot 10^2 + a_3 + a_4 \cdot 10 + a_5 \cdot 10^2 + ... + a_n \cdot 10^n \equiv 0 \text{ mod } 37 \Leftrightarrow 37 | z$\\
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\\
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Interpretiere dies als \enquote{Summe der Hunderter}:\\
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$(a_0+a_1 \cdot 10 + a_2 \cdot 10^2) + (a_3 + a_4 \cdot 10 + a_5 \cdot 10^2) + ... + a_n \cdot 10^n \equiv 0 \text{ mod } 37$\\
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\\
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$\Rightarrow \sum_{i=0}^{\lfloor \frac{n}{3} \rfloor}
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(a_{3i}+a_{3i+1}\cdot 10 + a_{3i+2} \cdot 10^2) \equiv 0 \text{ mod } 37$, aj = 0 für j>n\\
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$\Rightarrow$ das vorher beschriebene Kriterium
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\end{document}
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