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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 813 et 877 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 9 inférieur à 127 ?
        }

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

{\color{green!60!black}
        Quel est le plus petit multiple de 15 supérieur à 281 ?
        }

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

{\color{green!60!black}
\begin{enumerate}
\item Décompose 1088 et 684 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}
\(\dfrac{1088}{684}\)
\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  224 et 8625.
 \item Donne la liste ordonnée de leurs diviseurs communs.
 \item Déduis-en le PGCD de 224 et 8625.
 \item Simplifie au maximum la fraction :
\par\addvspace{.55em}
\(\dfrac{224}{8625}\)
\end{enumerate}
}

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

{\color{green!60!black} Les nombres suivants sont-ils premiers ? 1775;  367;  9092; 3057 }

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

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

{\color{green!60!black}
        Encadre 813 et 877 par deux multiples consécutifs de 6.
        }

On effectue la division euclidienne de 813 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) {1};
  \node[black, anchor=base, font=\ttfamily \fontsize{6.3}{7.5}\selectfont] at (1.540,1.980) {3};
  \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) {3};
  \node[blue, anchor=base, font=\ttfamily \fontsize{6.3}{7.5}\selectfont] at (2.640,1.650) {5};
  \node[red, anchor=base, font=\ttfamily \fontsize{6.3}{7.5}\selectfont] at (1.100,1.760) {6};
  \node[black, anchor=base, font=\ttfamily \fontsize{6.3}{7.5}\selectfont] at (1.320,1.540) {1};
  \node[green!55!black, anchor=base, font=\ttfamily \fontsize{6.3}{7.5}\selectfont] at (1.100,1.540) {2};
  \node[red, anchor=base, font=\ttfamily \fontsize{6.3}{7.5}\selectfont] at (1.320,1.320) {8};
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  \node[green!55!black, anchor=base, font=\ttfamily \fontsize{6.3}{7.5}\selectfont] at (1.540,0.605) {3};
\end{tikzpicture}\end{center}


\begin{itemize}
\item[] 813 = 6 \(\times\) 135 + 3 et 3 \(<\) 6
 \item[] 813 = 810 + 3
 \item[] {\color{red}
donc 810 \(<\) 813 \(<\) 816 (810 + 6)}
\end{itemize}

De même:

On effectue la division euclidienne de 877 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) {7};
  \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) {4};
  \node[blue, anchor=base, font=\ttfamily \fontsize{6.3}{7.5}\selectfont] at (2.640,1.650) {6};
  \node[red, anchor=base, font=\ttfamily \fontsize{6.3}{7.5}\selectfont] at (1.100,1.760) {6};
  \node[black, anchor=base, font=\ttfamily \fontsize{6.3}{7.5}\selectfont] at (1.320,1.540) {7};
  \node[green!55!black, anchor=base, font=\ttfamily \fontsize{6.3}{7.5}\selectfont] at (1.100,1.540) {2};
  \node[red, anchor=base, font=\ttfamily \fontsize{6.3}{7.5}\selectfont] at (1.320,1.320) {4};
  \node[red, anchor=base, font=\ttfamily \fontsize{6.3}{7.5}\selectfont] at (1.100,1.320) {2};
  \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) {3};
  \node[red, anchor=base, font=\ttfamily \fontsize{6.3}{7.5}\selectfont] at (1.540,0.880) {6};
  \node[red, anchor=base, font=\ttfamily \fontsize{6.3}{7.5}\selectfont] at (1.320,0.880) {3};
  \node[green!55!black, anchor=base, font=\ttfamily \fontsize{6.3}{7.5}\selectfont] at (1.540,0.605) {1};
\end{tikzpicture}\end{center}


\begin{itemize}
\item[] 877 = 6 \(\times\) 146 + 1 et 1 \(<\) 6
 \item[] 877 = 876 + 1
 \item[] {\color{red}
donc 876 \(<\) 877 \(<\) 882 (876 + 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 9 inférieur à 127 ?
           }

On effectue la division euclidienne de 127 par 9 :

 
\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);
  \draw[blue] (0.990,1.705) .. controls (1.137,1.778) and (1.283,1.778) .. (1.430,1.705);
  \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) {2};
  \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) {9};
  \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) {4};
  \node[red, anchor=base, font=\ttfamily \fontsize{6.3}{7.5}\selectfont] at (1.320,1.320) {9};
  \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) {3};
  \node[red, anchor=base, font=\ttfamily \fontsize{6.3}{7.5}\selectfont] at (1.540,0.880) {6};
  \node[red, anchor=base, font=\ttfamily \fontsize{6.3}{7.5}\selectfont] at (1.320,0.880) {3};
  \node[green!55!black, anchor=base, font=\ttfamily \fontsize{6.3}{7.5}\selectfont] at (1.540,0.605) {1};
\end{tikzpicture}\end{center}


\begin{itemize}
\item[] 127 = 9 \(\times\) 14 + 1 et 1 \(<\) 9
\item[] {\color{red} Le plus grand multiple de 9 inférieur à 127 est donc 9 \(\times\) 14 = 126.}
\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 15 supérieur à 281 ?
           }

On effectue la division euclidienne de 281 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] (1.045,0.825) -- (1.595,0.825);
  \draw[blue] (0.990,1.705) .. controls (1.137,1.778) and (1.283,1.778) .. (1.430,1.705);
  \node[black, anchor=base, font=\ttfamily \fontsize{6.3}{7.5}\selectfont] at (1.100,1.540) {2};
  \node[black, anchor=base, font=\ttfamily \fontsize{6.3}{7.5}\selectfont] at (1.320,1.540) {8};
  \node[black, anchor=base, font=\ttfamily \fontsize{6.3}{7.5}\selectfont] at (1.540,1.540) {1};
  \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) {8};
  \node[red, anchor=base, font=\ttfamily \fontsize{6.3}{7.5}\selectfont] at (1.320,1.320) {5};
  \node[red, anchor=base, font=\ttfamily \fontsize{6.3}{7.5}\selectfont] at (1.100,1.320) {1};
  \node[black, anchor=base, font=\ttfamily \fontsize{6.3}{7.5}\selectfont] at (1.540,1.100) {1};
  \node[green!55!black, anchor=base, font=\ttfamily \fontsize{6.3}{7.5}\selectfont] at (1.320,1.100) {3};
  \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 (1.540,0.880) {0};
  \node[red, anchor=base, font=\ttfamily \fontsize{6.3}{7.5}\selectfont] at (1.320,0.880) {2};
  \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 (1.540,0.605) {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[] 281 = 15 \(\times\) 18 + 11 et 11 \(<\) 15
\item[] {\color{red} Le plus petit multiple de 15 inférieur à 281 est donc 15 \(\times\) 19 = 285.}
\end{itemize}

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

{\color{green!60!black}
    Décomposition de  1088 en produit de facteurs premiers :

    
{\color{black}\begin{center}
\renewcommand{\arraystretch}{1.1}
\begin{tabular}{r|l}
1088 & 2\\
544 & 2\\
272 & 2\\
136 & 2\\
68 & 2\\
34 & 2\\
17 & 17\\
1 & \\
\end{tabular}\qquad 1088 = 2\textsuperscript{6} $\times$ 17
\end{center}}

    Décomposition de  684 en produit de facteurs premiers :

    
{\color{black}\begin{center}
\renewcommand{\arraystretch}{1.1}
\begin{tabular}{r|l}
684 & 2\\
342 & 2\\
171 & 3\\
57 & 3\\
19 & 19\\
1 & \\
\end{tabular}\qquad 684 = 2\textsuperscript{2} $\times$ 3\textsuperscript{2} $\times$ 19
\end{center}}

\begin{enumerate}
\item {\color{red} Décompositions :
\par\addvspace{.55em}
{\color{green!60!black} 1088 = 2\textsuperscript{6} \(\times\) 17}
\par\addvspace{.55em}
{\color{green!60!black} 684 = 2\textsuperscript{2} \(\times\) 3\textsuperscript{2} \(\times\) 19}}
\item {\color{red} Calcul du plus petit multiple commun :
\par\addvspace{.55em}
{\color{green!60!black} PPCM(1088;684) = 2\textsuperscript{6} \(\times\) 3\textsuperscript{2} \(\times\) 17 \(\times\) 19 = 186048}}
\item {\color{red} Calcul du plus grand
         diviseur commun :
\par\addvspace{.55em}
{\color{green!60!black} PGCD(1088;684) = 2\textsuperscript{2} = 4}}
\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 \(\dfrac{1088}{684}\) = \(\dfrac{1088:4}{684:4}\) = \(\dfrac{272}{171}\) } }
\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} 224 : \{ 1; 2; 4; 7; 8; 14; 16; 28; 32; 56; 112; 224 \}
\par\addvspace{.55em}

8625 : \{ 1; 3; 5; 15; 23; 25; 69; 75; 115; 125; 345; 375; 575; 1725; 2875; 8625 \}}

}
 \item {\color{red} Les diviseurs communs de 224 et 8625 sont :

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

 }
\item {\color{red} Le plus grand diviseur commun de 224 et 8625 est :

{\color{green!60!black}
PGCD(224;8625) = 1}

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

{\color{green!60!black} Leurs PGCD étant égal à 1, les nombres
224 et 8625 sont premiers entre eux.}

 {\color{green!60!black} Donc la fraction \(\dfrac{224}{8625}\) est irréductible.}}
\end{enumerate}
}

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

\begin{enumerate}
\item {\color{red}
        1775 est-il premier ?

{\color{green!60!black}

        1775 se termine par un 5, c'est un multiple de 5 donc \emph{1775 n'est pas un nombre premier}.}

}

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

{\color{green!60!black}  367 = 2 \(\times\) 183 + 1 367 = 3 \(\times\) 122 + 1 367 = 5 \(\times\) 73 + 2 367 = 7 \(\times\) 52 + 3 367 = 11 \(\times\) 33 + 4 367 = 13 \(\times\) 28 + 3 367 = 17 \(\times\) 21 + 10 367 = 19 \(\times\) 19 + 6 367 = 23 \(\times\) 15 + 22
\par\addvspace{.55em}

            367 n'a pas d'autres diviseurs que 1 et 367  donc \emph{ 367 est un nombre premier.} }

}
\item {\color{red}
        9092 est-il premier ?

{\color{green!60!black}

        9092 est pair donc \emph{9092 n'est pas un nombre premier}.}

}

\item {\color{red}
        3057 est-il premier ?

{\color{green!60!black}

          3+0+5+7 = 15
\par\addvspace{.55em}
 1+5 = 6

{\color{green!60!black} Critère de divisibilité par 3 :
\par\addvspace{.55em}

                     Un nombre entier est divisible par 3 si (et seulement si) la somme de ses chiffres est elle-même divisible par 3.
                     .}

                     donc 3057 est divisible par 3.\emph{ donc 3057 n'est pas un nombre premier}.}

}
\end{enumerate}
\end{document}
