Bunuel wrote:
What is the remainder when \(3^0 + 3^1 + 3^2 + ... + 3^{2009}\) is divided by 8?
(A) 0
(B) 1
(C) 2
(D) 4
(E) 6
slightly different approach and method i used
cyclicity of 3 ;
3^0=1 , 3^1 = 3, 3^2 = 9; 3^3= 7 ; 3^4 = 1
so for set of every 4 no we get a repeat of unit digits
1+3+9+7 ; 20 now 2009 times would be 20*100= 2000+ 9 units more i.e two more cycles of 3 ; 20+20 and 1 of 3^1= 43
our last digits would be ~ 2000+43; 2043
when 2043 divided by 8 ; 255*8 ; 2040 ; ~ 3 remainder ; best answer is 4
IMO D
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