Question

DAU COROANA VA ROG EU MULT​

Answer 1

$\it n=2+2^2+2^3+\ ...\ 2^{2013} =\\ \\ = (2+2^2+2^3)+(2^4+2^5+2^6)+\ ...\ +(2^{2011}+2^{2012}+\it 2^{2013})=\\ \\ =2(\underbrace{\it1+2+\it2^2}_{=7})+2^4(1+2+2^2)+\ ...\ +2^{2011}(1+2+2^2)=\\ \\ =7\cdot(2+2^4+\ ...\ +2^{2011})\in M_{7} \Rightarrow n\ \vdots\ 7$