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\begin{document}

\begin{center}
    \small Les 6 exercices d'arithmétique ci-dessous peuvent être réalisés facilement avec une calculatrice. Mais maîtrisez-vous les automatismes nécessaires pour vous en passer ? Division euclidienne, multiples, décomposition en facteurs premiers, PPCM, PGCD, fractions irréductibles et nombres premiers : un tour d'horizon complet du programme de cycle 4, utile pour réviser le DNB.
\end{center}
\vspace{0.5em}
\sloppy


{\color{blue}

« Les 6 exercices suivants peuvent être réalisés facilement avec une calculatrice. Mais maîtrisez-vous les automatismes nécessaires pour vous en passer ? »

}

\needspace{3\baselineskip}\par\smallskip\noindent{\color{red}\textbf{Exercice 1}}\par

{\color{green!60!black}
    Encadre 885 et 967 par deux multiples consécutifs de 6.
    }

\needspace{3\baselineskip}\par\smallskip\noindent{\color{red}\textbf{Exercice 2}}\par

{\color{green!60!black}
        Quel est le plus grand multiple de 15 inférieur à 177 ?
        }

\needspace{3\baselineskip}\par\smallskip\noindent{\color{red}\textbf{Exercice 3}}\par

{\color{green!60!black}
        Quel est le plus petit multiple de 3 supérieur à 46 ?
        }

\needspace{3\baselineskip}\par\smallskip\noindent{\color{red}\textbf{Exercice 4}}\par

{\color{green!60!black}
\begin{enumerate}
\item Décompose 384 et 7440 en produit de facteurs premiers.
 \item Déduis leur PPCM (Plus petit multiple commun)
 \item Déduis en leur PGCD (Plus grand diviseur commun)
 \item Simplifie au maximum la fraction :
\par\addvspace{.55em}
\(\dfrac{384}{7440}\)
\end{enumerate}
}

\needspace{3\baselineskip}\par\smallskip\noindent{\color{red}\textbf{Exercice 5}}\par

{\color{green!60!black}
\begin{enumerate}
\item Donne les listes ordonnées des diviseurs de  6561 et 544.
 \item Donne la liste ordonnée de leurs diviseurs communs.
 \item Déduis-en le PGCD de 6561 et 544.
 \item Simplifie au maximum la fraction :
\par\addvspace{.55em}
\(\dfrac{6561}{544}\)
\end{enumerate}
}

"
\needspace{3\baselineskip}\par\smallskip\noindent{\color{red}\textbf{Exercice 6}}\par

{\color{green!60!black} Les nombres suivants sont-ils premiers ? 653;  4807;  1005; 7204 }

 \newpage
\section*{\textcolor{red}{Correction}}

\needspace{3\baselineskip}\par\smallskip\noindent{\color{red}\textbf{Exercice 1}}\par

{\color{green!60!black}
        Encadre 885 et 967 par deux multiples consécutifs de 6.
        }

On effectue la division euclidienne de 885 par 6 :

 
\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[gray] (1.320,1.870) -- (1.320,1.760);
  \draw[gray] (1.540,1.870) -- (1.540,1.320);
  \draw[black] (1.980,2.145) -- (1.980,0.220);
  \draw[black] (1.980,1.870) -- (3.080,1.870);
  \draw[black] (1.045,1.705) -- (1.155,1.705);
  \draw[black] (1.045,1.265) -- (1.375,1.265);
  \draw[black] (1.265,0.825) -- (1.595,0.825);
  \draw[blue] (0.990,2.145) .. controls (1.063,2.218) and (1.137,2.218) .. (1.210,2.145);
  \node[black, anchor=base, font=\ttfamily \fontsize{6.3}{7.5}\selectfont] at (1.100,1.980) {8};
  \node[black, anchor=base, font=\ttfamily \fontsize{6.3}{7.5}\selectfont] at (1.320,1.980) {8};
  \node[black, anchor=base, font=\ttfamily \fontsize{6.3}{7.5}\selectfont] at (1.540,1.980) {5};
  \node[black, anchor=base west, font=\ttfamily \fontsize{6.3}{7.5}\selectfont] at (2.200,1.980) {6};
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  \node[green!55!black, anchor=base, font=\ttfamily \fontsize{6.3}{7.5}\selectfont] at (1.540,0.605) {3};
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\begin{itemize}
\item[] 885 = 6 \(\times\) 147 + 3 et 3 \(<\) 6
 \item[] 885 = 882 + 3
 \item[] {\color{red}
donc 882 \(<\) 885 \(<\) 888 (882 + 6)}
\end{itemize}

De même:

On effectue la division euclidienne de 967 par 6 :

 
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  \draw[gray] (1.320,1.870) -- (1.320,1.760);
  \draw[gray] (1.540,1.870) -- (1.540,1.320);
  \draw[black] (1.980,2.145) -- (1.980,0.220);
  \draw[black] (1.980,1.870) -- (3.080,1.870);
  \draw[black] (1.045,1.705) -- (1.155,1.705);
  \draw[black] (1.045,1.265) -- (1.375,1.265);
  \draw[black] (1.265,0.825) -- (1.595,0.825);
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  \node[black, anchor=base, font=\ttfamily \fontsize{6.3}{7.5}\selectfont] at (1.100,1.980) {9};
  \node[black, anchor=base, font=\ttfamily \fontsize{6.3}{7.5}\selectfont] at (1.320,1.980) {6};
  \node[black, anchor=base, font=\ttfamily \fontsize{6.3}{7.5}\selectfont] at (1.540,1.980) {7};
  \node[black, anchor=base west, font=\ttfamily \fontsize{6.3}{7.5}\selectfont] at (2.200,1.980) {6};
  \node[blue, anchor=base, font=\ttfamily \fontsize{6.3}{7.5}\selectfont] at (2.200,1.650) {1};
  \node[blue, anchor=base, font=\ttfamily \fontsize{6.3}{7.5}\selectfont] at (2.420,1.650) {6};
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\begin{itemize}
\item[] 967 = 6 \(\times\) 161 + 1 et 1 \(<\) 6
 \item[] 967 = 966 + 1
 \item[] {\color{red}
donc 966 \(<\) 967 \(<\) 972 (966 + 6)}
\end{itemize}

\needspace{3\baselineskip}\par\smallskip\noindent{\color{red}\textbf{Exercice 2}}\par

{\color{green!60!black}
           Quel est le plus grand multiple de 15 inférieur à 177 ?
           }

On effectue la division euclidienne de 177 par 15 :

 
\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[gray] (1.540,1.430) -- (1.540,1.320);
  \draw[black] (1.980,1.705) -- (1.980,0.220);
  \draw[black] (1.980,1.430) -- (2.860,1.430);
  \draw[black] (1.045,1.265) -- (1.375,1.265);
  \draw[black] (1.265,0.825) -- (1.595,0.825);
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  \node[black, anchor=base, font=\ttfamily \fontsize{6.3}{7.5}\selectfont] at (1.100,1.540) {1};
  \node[black, anchor=base, font=\ttfamily \fontsize{6.3}{7.5}\selectfont] at (1.320,1.540) {7};
  \node[black, anchor=base, font=\ttfamily \fontsize{6.3}{7.5}\selectfont] at (1.540,1.540) {7};
  \node[black, anchor=base west, font=\ttfamily \fontsize{6.3}{7.5}\selectfont] at (2.200,1.540) {15};
  \node[blue, anchor=base, font=\ttfamily \fontsize{6.3}{7.5}\selectfont] at (2.200,1.210) {1};
  \node[blue, anchor=base, font=\ttfamily \fontsize{6.3}{7.5}\selectfont] at (2.420,1.210) {1};
  \node[red, anchor=base, font=\ttfamily \fontsize{6.3}{7.5}\selectfont] at (1.320,1.320) {5};
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  \node[black, anchor=base, font=\ttfamily \fontsize{6.3}{7.5}\selectfont] at (1.540,1.100) {7};
  \node[green!55!black, anchor=base, font=\ttfamily \fontsize{6.3}{7.5}\selectfont] at (1.320,1.100) {2};
  \node[red, anchor=base, font=\ttfamily \fontsize{6.3}{7.5}\selectfont] at (1.540,0.880) {5};
  \node[red, anchor=base, font=\ttfamily \fontsize{6.3}{7.5}\selectfont] at (1.320,0.880) {1};
  \node[green!55!black, anchor=base, font=\ttfamily \fontsize{6.3}{7.5}\selectfont] at (1.540,0.605) {2};
  \node[green!55!black, anchor=base, font=\ttfamily \fontsize{6.3}{7.5}\selectfont] at (1.320,0.605) {1};
\end{tikzpicture}\end{center}


\begin{itemize}
\item[] 177 = 15 \(\times\) 11 + 12 et 12 \(<\) 15
\item[] {\color{red} Le plus grand multiple de 15 inférieur à 177 est donc 15 \(\times\) 11 = 165.}
\end{itemize}

\needspace{3\baselineskip}\par\smallskip\noindent{\color{red}\textbf{Exercice 3}}\par

{\color{green!60!black}
            Quel est le plus petit multiple de 3 supérieur à 46 ?
           }

On effectue la division euclidienne de 46 par 3 :

 
\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[gray] (1.320,1.430) -- (1.320,1.320);
  \draw[black] (1.760,1.705) -- (1.760,0.220);
  \draw[black] (1.760,1.430) -- (2.640,1.430);
  \draw[black] (1.045,1.265) -- (1.155,1.265);
  \draw[black] (1.045,0.825) -- (1.375,0.825);
  \draw[blue] (0.990,1.705) .. controls (1.063,1.778) and (1.137,1.778) .. (1.210,1.705);
  \node[black, anchor=base, font=\ttfamily \fontsize{6.3}{7.5}\selectfont] at (1.100,1.540) {4};
  \node[black, anchor=base, font=\ttfamily \fontsize{6.3}{7.5}\selectfont] at (1.320,1.540) {6};
  \node[black, anchor=base west, font=\ttfamily \fontsize{6.3}{7.5}\selectfont] at (1.980,1.540) {3};
  \node[blue, anchor=base, font=\ttfamily \fontsize{6.3}{7.5}\selectfont] at (1.980,1.210) {1};
  \node[blue, anchor=base, font=\ttfamily \fontsize{6.3}{7.5}\selectfont] at (2.200,1.210) {5};
  \node[red, anchor=base, font=\ttfamily \fontsize{6.3}{7.5}\selectfont] at (1.100,1.320) {3};
  \node[black, anchor=base, font=\ttfamily \fontsize{6.3}{7.5}\selectfont] at (1.320,1.100) {6};
  \node[green!55!black, anchor=base, font=\ttfamily \fontsize{6.3}{7.5}\selectfont] at (1.100,1.100) {1};
  \node[red, anchor=base, font=\ttfamily \fontsize{6.3}{7.5}\selectfont] at (1.320,0.880) {5};
  \node[red, anchor=base, font=\ttfamily \fontsize{6.3}{7.5}\selectfont] at (1.100,0.880) {1};
  \node[green!55!black, anchor=base, font=\ttfamily \fontsize{6.3}{7.5}\selectfont] at (1.320,0.605) {1};
\end{tikzpicture}\end{center}


\begin{itemize}
\item[] 46 = 3 \(\times\) 15 + 1 et 1 \(<\) 3
\item[] {\color{red} Le plus petit multiple de 3 inférieur à 46 est donc 3 \(\times\) 16 = 48.}
\end{itemize}

\needspace{3\baselineskip}\par\smallskip\noindent{\color{red}\textbf{Exercice 4}}\par

{\color{green!60!black}
    Décomposition de  384 en produit de facteurs premiers :

    
{\color{black}\begin{center}
\renewcommand{\arraystretch}{1.1}
\begin{tabular}{r|l}
384 & 2\\
192 & 2\\
96 & 2\\
48 & 2\\
24 & 2\\
12 & 2\\
6 & 2\\
3 & 3\\
1 & \\
\end{tabular}\qquad 384 = 2\textsuperscript{7} $\times$ 3
\end{center}}

    Décomposition de  7440 en produit de facteurs premiers :

    
{\color{black}\begin{center}
\renewcommand{\arraystretch}{1.1}
\begin{tabular}{r|l}
7440 & 2\\
3720 & 2\\
1860 & 2\\
930 & 2\\
465 & 3\\
155 & 5\\
31 & 31\\
1 & \\
\end{tabular}\qquad 7440 = 2\textsuperscript{4} $\times$ 3 $\times$ 5 $\times$ 31
\end{center}}

\begin{enumerate}
\item {\color{red} Décompositions :
\par\addvspace{.55em}
{\color{green!60!black} 384 = 2\textsuperscript{7} \(\times\) 3}
\par\addvspace{.55em}
{\color{green!60!black} 7440 = 2\textsuperscript{4} \(\times\) 3 \(\times\) 5 \(\times\) 31}}
\item {\color{red} Calcul du plus petit multiple commun :
\par\addvspace{.55em}
{\color{green!60!black} PPCM(384;7440) = 2\textsuperscript{7} \(\times\) 3 \(\times\) 5 \(\times\) 31 = 59520}}
\item {\color{red} Calcul du plus grand
         diviseur commun :
\par\addvspace{.55em}
{\color{green!60!black} PGCD(384;7440) = 2\textsuperscript{4} \(\times\) 3 = 48}}
\item {\color{red} Simplification de la fraction :
\par\addvspace{.55em}

{\color{green!60!black} Si on simplifie une fraction par
         le plus grand diviseur commun à son numérateur
         et son dénominateur, la fraction obtenue est irréductible.}

 {\color{green!60!black} d'où \(\dfrac{384}{7440}\) = \(\dfrac{384:48}{7440:48}\) = \(\dfrac{8}{155}\) } }
\end{enumerate}
}

\needspace{3\baselineskip}\par\smallskip\noindent{\color{red}\textbf{Exercice 5}}\par

{\color{green!60!black}
\begin{enumerate}
\item {\color{red} Les diviseurs :

{\color{green!60!black} 6561 : \{ 1; 3; 9; 27; 81; 243; 729; 2187; 6561 \}
\par\addvspace{.55em}

544 : \{ 1; 2; 4; 8; 16; 17; 32; 34; 68; 136; 272; 544 \}}

}
 \item {\color{red} Les diviseurs communs de 6561 et 544 sont :

{\color{green!60!black}
\{ 1 \}}

 }
\item {\color{red} Le plus grand diviseur commun de 6561 et 544 est :

{\color{green!60!black}
PGCD(6561;544) = 1}

}
 \item {\color{red} Simplification de la fraction :

{\color{green!60!black} Leurs PGCD étant égal à 1, les nombres
6561 et 544 sont premiers entre eux.}

 {\color{green!60!black} Donc la fraction \(\dfrac{6561}{544}\) est irréductible.}}
\end{enumerate}
}

\needspace{3\baselineskip}\par\smallskip\noindent{\color{red}\textbf{Exercice 6}}\par

\begin{enumerate}
\item {\color{red}
        653 est-il premier ?

{\color{green!60!black}  653 = 2 \(\times\) 326 + 1 653 = 3 \(\times\) 217 + 2 653 = 5 \(\times\) 130 + 3 653 = 7 \(\times\) 93 + 2 653 = 11 \(\times\) 59 + 4 653 = 13 \(\times\) 50 + 3 653 = 17 \(\times\) 38 + 7 653 = 19 \(\times\) 34 + 7 653 = 23 \(\times\) 28 + 9 653 = 29 \(\times\) 22 + 15
\par\addvspace{.55em}

            653 n'a pas d'autres diviseurs que 1 et 653  donc \emph{ 653 est un nombre premier.} }

}
\item {\color{red}
        4807 est-il premier ?

{\color{green!60!black}  4807 = 2 \(\times\) 2403 + 1 4807 = 3 \(\times\) 1602 + 1 4807 = 5 \(\times\) 961 + 2 4807 = 7 \(\times\) 686 + 5 4807 = 11 \(\times\) 437 + 0
\par\addvspace{.55em}

            4807 est divisible par 11 donc\emph{  4807 n'est pas un nombre premier}.
            }

}
\item {\color{red}
        1005 est-il premier ?

{\color{green!60!black}

        1005 se termine par un 5, c'est un multiple de 5 donc \emph{1005 n'est pas un nombre premier}.}

}

\item {\color{red}
        7204 est-il premier ?

{\color{green!60!black}

        7204 est pair donc \emph{7204 n'est pas un nombre premier}.}

}

\end{enumerate}
\end{document}
