Question Number 104093 by bramlex last updated on 19/Jul/20

(1)Evaluate the sum ⌊(2^0 /3)⌋+⌊(2^1 /3)⌋+⌊(2^2 /3)⌋+...+⌊(2^(1000) /3)⌋  (2) find 2^(98)  (mod 33)

Answered by JDamian last updated on 19/Jul/20

(2) 2^(98) mod 33=[(2^5 )^(19) 2^3 ]mod 33=  ={[(2^5 )^(19) ]mod 33} ∙ (8 mod 33)=  ={[(32)^(19) ]mod 33} ∙ (8 mod 33)=  =(32 mod 33)^(19)  ∙ (8 mod 33)=  =(32 mod 33)^(19)  ∙ (8 mod 33)=  =[(−1)^(19)  ∙ 8] mod 33 = (−8)mod 33=  =(33−8)mod 33 = 25 mod 33 = 25

Answered by john santu last updated on 19/Jul/20

(1) Note that we have 2^x =   { ((1 (mod 3) , if x is even )),((2  (mod 3), if x is odd )) :}  Therefore we get Σ_(n = 0) ^(1000) ⌊(2^n /3)⌋ = 0  + Σ_(n = 1) ^(500) (⌊(2^(2n−1) /3)⌋+⌊(2^(2n) /3)⌋) = Σ_(n = 1) ^(500) (((2^(2n−1) −2)/3)+((2^(2n) −1)/3))  = (1/3)Σ_(n = 1) ^(500) (2^(2n−1) +2^(2n) −1)  = (1/3)Σ_(n = 1) ^(1000) 2^n −500   = (1/3)(2^(1001) −2)−500 .(JS ⊛)