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\begin{document}
\noindent
\begin{center}
    \begin{tabular}{@{}p{0.22\textwidth}p{0.57\textwidth}p{0.17\textwidth}}
        %\toprule
            \multirow{2}{*}{\parbox{\linewidth}{Prof MOSAID \newline \mylink }}
            & \Centering {Correction Control N 01 - A} & \hfill  TCS \\
        \bottomrule
    \end{tabular}
\end{center}
\noindent
\textbf{\underline{Exercice 1:}}\\
\noindent
Soient $x=2420$ et $y=6930$\\
1.a la décomposition:\\
\vspace*{-2cm}
\begin{center}
    \begin{tabular}{r|l}
        2420&2\\
        1210&2\\
        605&5\\
        121&11\\
        11&11\\
        1&\\
    \end{tabular}
    \hspace*{1cm}
    \begin{tabular}{r|l}
        6930&2\\
        3465&5\\
        693&3\\
        231&3\\
        77&7\\
        11&11\\
        1&\\
    \end{tabular}
\end{center}
Donc $x=2^2 \times 5 \times 11^2$ ~~~~et~~~~ $y=2 \times 3^2 \times 5 \times 7 \times 11$\\
On a $pgcd(x,y)=2 \times 5 \times 11 = 110$ ~~~et~~~ $ppcm(x,y)=2^2 \times 3^2 \times 5 \times 7 \times 11^2$\\
On a $\sqrt{\textcolor{red}{5}x}=\sqrt{\textcolor{red}{5}\times 2^2 \times 5\times 11^2}=
\sqrt{5^2\times 2^2 \times 11^2}=2 \times 5 \times 11 = 110$\\
$\sqrt{\frac{\textcolor{red}{35}y}{\textcolor{red}{22}}}=
\sqrt{\frac{\textcolor{red}{5 \times 7} \times 2 \times 3^2 \times 5 \times 7 \times 11 }{\textcolor{red}{2 \times 11}}}=
\sqrt{3^2 \times 5^2 \times 7^2 \times  11^2} = 3 \times 5 \times 7 \times 11$\\
Utiliser l'algorithme d'euclid pour determiner $pgcd(x,y)$\\
On a
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\begin{align*}
    6930&=2420 \times 2 + \textbf{2090}\\
    2420&=2090 \times 1 + \textbf{330}\\
    2090&=330 \times 6 + \textbf{\textcolor{red}{110}}\\
    330&=110 \times 3 + \textbf{0}
\end{align*}
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Le dernier rest non  nul est le pgcd donc  $pgcd(x,y)=110$\\
2. On a $\sqrt371=19.2$. les nombres premiers inférieurs ou égals à $\sqrt371$ sont 2,3,5,7,11,13,17 et 19\\
On a 371 n'est pas divisible par 2 car impaire\\
on a $3+7+1=11$ donc il n'est pas divisible par 3\\
aussi il n'est pas divisible par 5\\
On effectue la division euclidienne par 7 on trouve $371=7 \times 53$\\
Donc 371 n'est pas premier.\\
\stamp{14}{2}
\clearpage
3. Soint $ n \in\mathbb{N}$. On a $a=n^2-n+1$\\
\begin{multicols}{2}

si $n$ est paire \\
donc il existe $k \in\mathbb{N}$ tel que $n=2k$\\
donc
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\begin{align*}
    a&=n^2-n+1\\
    &=(2k)^2-2k+1\\
    &=2(2k^2-k)+1
\end{align*}
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donc impaire
\columnbreak

si $n$ est impaire\\
donc il existe $k \in\mathbb{N}$ tel que $n=2k+1$\\
donc
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\begin{align*}
    a&=n^2-n+1\\
    &=(2k+1)^2-(2k+1)+1\\
    &=4k^2+4k+1-2k+1-1\\
    &=4k^2+2k+1\\
    &=2(k^2+k)+1
\end{align*}
donc impaire
\end{multicols}
4. On a $D_{42}=\{1,42,2,21,3,14,4,6,7\}$\\
\noindent
\textbf{\underline{Exercice 2:}}\\
1. La figure:\\
$\overrightarrow{AI}=\frac{1}{2}\overrightarrow{AB}$\\
$\overrightarrow{AJ}=\frac{2}{5}\overrightarrow{AC}$\\
$\overrightarrow{BK}=-2\overrightarrow{BC}$\\
\hspace*{7cm}
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\begin{multicols}{2}
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On a:
\begin{align*}
    \overrightarrow{AJ}&=\frac{2}{5}\overrightarrow{AC}\\
    \overrightarrow{A\textcolor{red}{I}}+\overrightarrow{\textcolor{red}{I}J}&=\frac{2}{5}\overrightarrow{AC}\\
    \overrightarrow{IJ}&=-\overrightarrow{AI}+\frac{2}{5}\overrightarrow{AC}\\
    \overrightarrow{IJ}&=-\frac{1}{2}\overrightarrow{AB}+\frac{2}{5}\overrightarrow{AC}
\end{align*}
\hrule
\noindent
On a $\overrightarrow{IK}=\frac{5}{2}\overrightarrow{AB}-2\overrightarrow{AC}$\\
Donc  $\overrightarrow{IK}=-5\left(-\frac{1}{2}\overrightarrow{AB}+\frac{2}{5}\overrightarrow{AC}\right)$\\[0.5cm]
Alors  $\overrightarrow{IK}=-5\overrightarrow{IJ}$ \\
Donc les vecteurs sont colinèaires\\
Donc les points I,J et K sont alignés.
\columnbreak
\begin{align*}
    \overrightarrow{BK}&=-2\overrightarrow{BC}\\
    \overrightarrow{B\textcolor{red}{A}}+\overrightarrow{\textcolor{red}{A}I}
    +\overrightarrow{\textcolor{red}{I}K}&=-2\overrightarrow{B\textcolor{red}{A}}-2\overrightarrow{\textcolor{red}{A}C}\\
    \overrightarrow{\textcolor{red}{A}I}+\overrightarrow{\textcolor{red}{I}K}&=-3\overrightarrow{BA}-2\overrightarrow{AC}\\
    \frac{1}{2}\overrightarrow{AB}+\overrightarrow{IK}&=-3\overrightarrow{BA}-2\overrightarrow{AC}\\
    \overrightarrow{IK}&=-\frac{1}{2}\overrightarrow{AB}-3\overrightarrow{BA}-2\overrightarrow{AC}\\
    \overrightarrow{IK}&=-\frac{1}{2}\overrightarrow{AB}+3\overrightarrow{AB}-2\overrightarrow{AC}\\
    \overrightarrow{IK}&=-\frac{1}{2}\overrightarrow{AB}+\frac{6}{2}\overrightarrow{AB}-2\overrightarrow{AC}\\
    \overrightarrow{IK}&=\frac{5}{2}\overrightarrow{AB}-2\overrightarrow{AC}\\
\end{align*}
\textcolor{white}{.}\hfill \underline{MOSAID le \today}\\
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