Question

Demonstrati ca numarul a=(2n+1)(5n-3)(4n+1) este divizibil cu 3 oricare ar fi n apartine lui N.

Answer 1

Resturile unui numar la 3 pot fi 0,1,2 .Asadar avem 3 cazuri de analizat:

i) n=3k, k∈N :
a=(2*3k+1)(5*3k-3)(4*3k+1)
a=(6k+1)(15k-3)(12k+1)
a=3 (6k+1)(5k-1)(12k+1) : 3

ii)n=3k+1 :
a=[2(3k+1)+1] * [5(3k+1)-3] * [4(3k+1)+1]
a=(6k+3)(15k+2)(12k+5)
a=3 (2k+1)(15k+2)(12k+5) :3

iii)n=3k+2:
a=[2(3k+2)+1] * [5(3k+2)-3] * [4(3k+2)+1]
a=(6k+5)(15k+7)(12k+9)
a=3 (6k+9)(15k+7)(4k+3) :3

Deci a:3, ∀n∈N