Question

se da numarul a=1+3+3^2+3^3+...+3^2007. a). Aflati ultimele 2 cifre ale numarului a

Answer 1

$a=1+3+3^2+3^3+...+3^{2005}+3^{2006}+3^{2007} \\ \\ a = (1+3^2+3+3^3)+(3^4+3^6+3^5+3^7)+...+\\+(3^{2004}+3^{2006}+3^{2005}+3^{2007}) \\ \\ a = 1\cdot 10+3\cdot 10+3^4\cdot 10+3^5\cdot 10+...+3^{2004}\cdot 10+3^{2005}\cdot 10\\ a = 10\cdot (1+3+3^4+3^5+...+3^{2004}+3^{2005}) \\ \\ \Rightarrow \text{Ultima cifra este 0, acum trebuie sa aflam penultima cifra} \\ \\ U(1+3+3^4+3^5+...+3^{2004}+3^{2005}) = \\ \\ =U(1+3+1+3+1+3+...+1+3) \\ \text{Avem puterile 0,4,8,...,2004 } \rightarrow 4\cdot (0,1,2,3,...501) \\ \\ =U(1\cdot 502+3\cdot 502) = U(502\cdot 4) = 8$

=> Penultima cifra e 8, iar ultima cifra e 0.

=> ultimele 2 cifre sunt 8 si 0.