[tex] n\in \mathbb_{N}$ $\\ \\ 5n!+(n+1)! = 40(n-1)! \\ 5\cdot (n-1)!\cdot n + (n-1)!\cdot n\cdot (n+1) = 40(n-1)! \\ (n-1)! \cdot \Big(5n+n(n+1)\Big) = 40(n-1)! \Big | : (n-1)! \\ \\ $(putem imparti cu factorial, nu vom pierde solutii deoarece factorialul\\este diferit de 0)\\ \\ $ 5n+n(n+1) = 40 \Rightarrow n(5+n+1) = 40 \Rightarrow n(n+6) = 40 \\ \\ 4\cdot(4+6) = 40 \Rightarrow \boxed{n=4}[/tex]
