Question

nr 247,297,347 impartite la acelasi nr n,dau resturile 7,9,11.aflati n

Answer 1

247=n·c1+7;
297=n·c2+9;
347=n·c3+11;
⇒n·c1=247-7⇒n·c1=240;
⇒n·c2=297-9⇒n·c2=288;
⇒n·c3=347-11⇒n·c3=336;
⇒c.m.m.d.c[240,288,336]=2⁴·3=16·3=48;
⇒n=48;