Question

DAU COROANA SI PUNCTE AM NEVOIE RAPID VA ROG EU MULT​

Answer 1

Răspuns:

Explicație pas cu pas:

Answer 2

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