Question

Determinai ultima cifra a numerelor:a)2+2^2+2^3+...+2^45=

Answer 1

2
2^2=4
2^3=8
2^4=16
2^5=32

observam ca ultima cifra se repeta din 4 in 4 puteri
44:4=11   avem 11 cicluri 

2+4+6+8=20, deci ultima cifra este 0 la un ciclu
11*20 rezulta ultima cifra 0
 deci de la 2 +.....2^44 ultima cifra este 0.
2^45 are ultima cifra 2
0+2=2 deci ultima cifra a sumei este 2

Answer 2

[tex]S=2+2^{2}+2^{3}+...+2^{45}|*2\\2S=2*(2+2^{2}+2^{3}+...+2^{45})\\ 2S=2*2+2*2^{2}+2*2^{3}+...+2*2^{45}\\ 2S=2^{1+1}+2^{1+2}+2^{1+3}+...+2^{1+45}\\2S=2^{2}+2^{3}+2^{4}+...+2^{46}\\ 2S=(2+2^{2}+2^{3}+...+2^{45})+2^{46}-2\\2S=S+2^{46}-2\\2S-S=2^{46}-2\\ S=2^{46}-2\\Aflam\ ultima\ cifra\ lui\ 2^{46}\\2^{1}=2\\2^{2}=4\\2^{3}=8\\ 2^{4}=16\\2^{5}=32\\2^{6}=64\\Puterile\ lui\ 2\ se\ termina\ in:2,4,8,6\\ 46:4=11\ rest\ 2\\U(2^{46})=U(2^{4*11+2})=U(2^{2})=4\\ U(2^{46}-2)=U(4-2)=U(2)=2\\Ultima\ cifra\ este\ 2[/tex]