[tex] 2^{n+3}\cdot 5^n + 1 = (2\cdot 5)^n\cdot 2^3 + 1 = 10^n\cdot 8 + 1 \\ \\ = \overline{\underbrace{8000\cdots 0001}_\text{n+1 cifre}}\text{(daca n = 0 avem 8 + 1 = 9)}\\ \\ \text{Suma cifre} = 8 + 1 = 9 \implies 3\mid 2^{n+3}\cdot 5^n + 1, \forall n \in \mathbb{N}[/tex]
