1. fie S=n+(n+1)+(n+2)+(n+3)+(n+4)+(n+5)+(n+6)+(n+7)+(n+8)+(n+9)=(10n +45) suma tuturor numerelor
10n +45 -(n+x) = 106 9n +45-x= 106 9n-x=61 ⇒ 9n-x=63-2 ⇒n=7; x=2 ( am ales 63 ptr.ca e divizibil cu 9, n∈N)
S=10×7 + 45 = 115 115-(7+2) = 106
2. x=1006(1007-2)-1005²=1006·1005 -1005²=1005(1006-1005) =1005⇒ S=1+0+0+5 = 6
3. 3n+n-1 =303 + 100 4n= 404 ⇒ n=101 ⇒ P=101×100=10100