Question

Detrminati ultima cifra a numerelor: b=7+7²+7³+...+7^45 c=3^0+3¹+3²+3³+....3^42 d=8+8²+8³+....8^88 DAU CROANA!!! MULTUMESC!!!

Answer 1

[tex]b=7+7^{2}+7^{3}+...+7^{45}|*7\\7b=7*(7+7^{2}+7^{3}+...+7^{45})\\ 7b=7*7+7*7^{2}+7*7^{3}+...+7*7^{45}\\7b=7^{1+1}+7^{1+2}+7^{1+3}+....+7^{1+45}\\7b=7^{2}+7^{3}+7^{4}+...+7^{46}\\ 7b=(7+7^{2}+7^{3}+...+7^{45})+7^{46}-7\\7b=b+7^{46}-7\\7b-b=7^{46}-7\\ 6b=7^{46}-7\\b=\frac{7^{46}-7}{6}\\7^{1}=7\\7^{2}=49\\7^{3}=343\\ 7^{4}=2401\\7^{5}=16807\\Puterile\ lui\ 7\ se\ termina\ in: 7,9,3,1\\ Sunt\ seturi\ de\ 4.\\U(7^{46})=U(7^{4*11+2})=U(7^{2})=U(49)\\ \frac{49-7}{6}=\frac{42}{6}=7\\Ultima\ cifra\ a\ lui\ b=7[/tex]
Celelalte exercitii se rezolva la fel.