Question

Fie n apartine N.Calculati S=(n+1;2n+5)+(n+2;2n+7)+(n+1002;2n+2007),unde cu (a,b) an notat cel mai mare divizor comun al numerelor naturale a si b.

Answer 1

Luam fiecare paranteza in parte:
n+1/2n+5  =>n+1/2n+5
n+1/n+1   =>n+1/2n+2
Le scadem: n+1/3=>(n+3,2n+5)=3
n+2/2n+7=>n+2/2n+7
n+2/n+2  =>n+2/2n+4
  Le scadem:n+2/3=>(n+2,2n+7)=3
n+1002/2n+2007=>n+1002/2n+2007
n+1002/n+1002 => n+1002/2n+2004
  Le scadem:        n+1002/3=>(n+1002,2n+2007)=3
S=3+3+3
S=9