Question

Aflati x,y apartin lui Z, stiind ca 33x+26y=715

Answer 1

$33x ~ \vdots ~11~si~715~ \vdots~11 \Rightarrow 26y~ \vdots~11,~si~cum~(26,11)=1 \Rightarrow y ~ \vdots~11. \\ \\Deci~y=11b,~unde~b \in Z. \\ \\ Ecuatia~devine:~3 \cdot 11x+26 \cdot 11b=11 \cdot 65 \Leftrightarrow 3x+26b=65. \\ \\ 26b ~ \vdots~13~si~65 ~\vdots ~13 \Rightarrow 3x ~ \vdots~13,~si~cum~(3,13)=1 \Rightarrow x~ \vdots ~13 . \\ \\ Deci~x=13a. \\ \\ Ecuatia~devine:~3 \cdot 13a+2 \cdot 13b=5 \cdot 13 \Leftrightarrow 3a+2b=5.$

$Din~ultima~relatie~rezulta~a= \frac{5-2b}{3}~si~b= \frac{5-3a}{2}. \\ \\ b \in Z \Rightarrow 2|5-3a \Rightarrow 3a=impar \Rightarrow a=2m+1~cu~m \in Z. \\ \\ a \in Z \Rightarrow 3 ~|~5-2b .~dar~3~|~6-3b \Rightarrow 3 ~|~(5-2b)-(6-3b) \Leftrightarrow 3~|~b-1.\\ \\Deci~b=3n+1,~cu~ n \in Z. \\ \\ a= \frac{5-6n-2}{3}= \frac{3-6n}{3}=1-2n~si~b= \frac{5-6m-3}{2}= \frac{2-6m}{2}=1-3m. \\ \\ Avem~deci~2m+1=1-2n~si~3n+1=1-3m.~Rezulta~m= -n. \\ \\ Asadar~a=1+2m~si~b=1-3m.$

$x=13a=13(1+2m)=13+26m. \\ \\ y=11(1-3m)=11-33m. \\ \\ Solutia~generala~a~ecuatiei~este~(x;y)=(13+26m;11-33m). \\ \\ *Aceasta~este~o~ecuatie~diofantica~de~gradul~1~cu~doua~necunoscute,~ \\ \\ de~aceea~admite~o~infinitate~de~solutii~intregi.$