Aratati ca 17 divide (3a+6b+8c) daca si numai daca 17 divide (11a+5b+c)

[tex]\boxed{\Rightarrow}~Demonstram~ca~daca~17~|~(3a+6b+8c),~atunci~ \\ 17~|~(11a+5b+c). \\ ------------------------------ \\ \\ 17~|~3a+6b+8c \Rightarrow 17~|~2(3a+6b+8c) \Leftrightarrow 17 ~|~6a+12b+16c.~~~~~~(1). \\ \\ 17~|~17(a+b+c)~\Leftrightarrow 17 ~|~17a+17b+17c.~~~~~~(2). \\ \\ Din~(1)~si~(2),~rezulta~17~|~(17a+17b+17c)-(6a+12b+16c) \Leftrightarrow \\ \ \\ \Leftrightarrow \boxed{17~|~11a+5b+c}.[/tex]

[tex]\boxed{ \Leftarrow }~Demonstram~ca~daca~17~|~(11a+5b+c),~atunci~ \\ 17~|~(3a+6b+8c). \\ ------------------------------ \\ \\ 17~|~17(a+b+c)~\Leftirghtarrow~17~|~17a+17b+17c. \\ \\ Dar~17~|~11a+5b+c~\Rightarrow~17~|~(17a+17b+17c)-(11a+5b+c) \Leftrightarrow \\ \\ \Leftrightarrow 17~|~6a+12b+16c~\Letrightarrow~17~|~2(3a+6b+8c),~dar~(17;2)=1~\Rightarrow \\ \\ \Rightarrow~\boxed{17~|~3a+6b+8c}.[/tex]

Folosim o proprietate a divizibilitatii : Daca intr-o suma de doi termeni, suma si unul din termeni se divid cu acelasi numar natural atunci si celalalt termen se divide cu acel numar natural
Notam A=3a+6b+8c
B=11a+5b+c
Avem  2A+B=17(a+b+c)
a) Daca 17|A cum 17|17(a+b+c)=> 17|B
b) Daca 17|B cum 17|17(a+b+c)=>17|A