Question

determinati nr naturale nenule cuprinse intre 900 si 1000 care impartite la 6 dau restul 5 impartite la 7 dau restul 6 si la 11 dau restul 10

Answer 1

x:6=a,r=5   ⇒x=6a+5  ⇒x+1=6a+6⇒x+1=6(a+1)
x:7=b,r=6   ⇒ x=7b+6  ⇒x+1=7b+7 ⇒x+1=7(b+1)
x:11=c,r=10⇒x=11c+10⇒x+1=11c+11⇒x+1=11(c+1)
Din amantrei rezulta:
x+1=[6,7,11]
x+1=462k
900<x<1000
900<462k<1000
1 intreg 438/462<k<2 intreg 76/462
         k∈N⇒k=2
x+1=462*2
x+1=924
x=923