N=2+2^2+2^3+...+2^120= avem 120 termeni deci putem sa-i grupam cate 4 astfel: =(2+2^2+2^3+2^4)+...+(2^117+2^118+2^119+2^120)= =2(1+2+4+8)+....+2^117(1+2+4+8)= =2*15+....+2^117 *15= =15*(2+...+2^117)= =3*5*(2+...+2^117) deci se divide cu numarul 3 şi 5