Question

3 la puterea 2018- 2 la puterea 2018 este divizibil cu 5

Answer 1

[tex]3^{2018}-2^{2018}=(3^2)^{1009}-(2^2)^{1009}=9^{1009}-4^{1009}\\ \text{Mai departe folosim uramtoarea proprietate:}\\ \boxed{a^n-b^n\ \vdots\ a-b,\forall n,a,b\in \mathbb{N}}\\ \text{Asadar:} 9^{1009}-4^{1009}\ \vdots 9-4 ,\text{adica }9^{1009}-4^{1009}\vdots 5[/tex]