Question

Demonstrati ca numarul a = $6^{n+2}$ + 5 · $6^{n}$ + $6^{n+1}$ se divide cu numarul 47, oricare ar fi exponentul n ∈ N .

Answer 1

[tex]a=6^{n+2}+5\cdot 6^n+6^{n+1}\\ a=6^n\cdot6^2+5\cdot 6^n+6^n\cdot6\\ a=6^n(36+6+5)\\ a=6^n\cdot 47\\ Deci\ a\vdots 47\ \forall n\in N.[/tex]