[tex]12 = 3 \times 4 \\
=\ \textgreater \ \texttt{Numarul }~ \overline{25ab0}~ \texttt{ trebuie sa fie divizibil si cu 3 si cu 4. } \\
\texttt{Il alegem pe b, astfel incat } ~ \overline{25ab0}~\texttt{ sa fie divizibil cu 4. } \\
\Longrightarrow~~b \in \{0;~2;~4;~8\} \\
\texttt{Il alegem pe a, astfel incat } ~ \overline{25ab0}~\texttt{ sa fie divizibil cu 3. }
[/tex]
[tex]\texttt{Daca b = 0, atunci } ~a \in \{2;~5;~8\} \\
\texttt{Daca b = 2 atunci } ~a \in \{0;~3;~6;~9\} \\
\texttt{Daca b = 4 atunci } ~a \in \{1;~4;~7\} \\
\texttt{Daca b = 6 atunci } ~a \in \{2;~5;~8\} \\
\texttt{Daca b = 8 atunci } ~a \in \{0;~3;~6;~9\} [/tex]
[tex]b) \displaystyle \\
Daca (x+1)|(2x+10) \Longrightarrow ~~fractia ~~ \frac{2x+10}{x+1} \in Z \\ \\
\frac{2x+10}{x+1} = \frac{2x+2+8}{x+1} = \frac{2x+2}{x+1} +\frac{8}{x+1} = 2 + \frac{8}{x+1} \\ \\
D_8 = \{-8;~-4;~-2;~-1;~1;~2;~4;~8\} \\
x + 1 = -8 ~~ =\ \textgreater \ ~~ x = -9 \\
x + 1 = -4 ~~ =\ \textgreater \ ~~ x = -5 \\
x + 1 = -2 ~~ =\ \textgreater \ ~~ x = -3 \\
x + 1 = -1 ~~ =\ \textgreater \ ~~ x = -2 \\
x + 1 = 1 ~~ =\ \textgreater \ ~~ x = 0 \\
x + 1 = 2 ~~ =\ \textgreater \ ~~ x = 1 \\
x + 1 = 4 ~~ =\ \textgreater \ ~~ x = 3 \\
x + 1 = 8 ~~=\ \textgreater \ ~~ x = 7 [/tex]
