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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)

$$\left(\mathrm{1}\right){Evaluate}\:{the}\:{sum}\:\lfloor\frac{\mathrm{2}^{\mathrm{0}} }{\mathrm{3}}\rfloor+\lfloor\frac{\mathrm{2}^{\mathrm{1}} }{\mathrm{3}}\rfloor+\lfloor\frac{\mathrm{2}^{\mathrm{2}} }{\mathrm{3}}\rfloor+...+\lfloor\frac{\mathrm{2}^{\mathrm{1000}} }{\mathrm{3}}\rfloor \\ $$$$\left(\mathrm{2}\right)\:{find}\:\mathrm{2}^{\mathrm{98}} \:\left({mod}\:\mathrm{33}\right)\: \\ $$

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

$$\left(\mathrm{2}\right)\:\mathrm{2}^{\mathrm{98}} {mod}\:\mathrm{33}=\left[\left(\mathrm{2}^{\mathrm{5}} \right)^{\mathrm{19}} \mathrm{2}^{\mathrm{3}} \right]{mod}\:\mathrm{33}= \\ $$$$=\left\{\left[\left(\mathrm{2}^{\mathrm{5}} \right)^{\mathrm{19}} \right]{mod}\:\mathrm{33}\right\}\:\centerdot\:\left(\mathrm{8}\:{mod}\:\mathrm{33}\right)= \\ $$$$=\left\{\left[\left(\mathrm{32}\right)^{\mathrm{19}} \right]{mod}\:\mathrm{33}\right\}\:\centerdot\:\left(\mathrm{8}\:{mod}\:\mathrm{33}\right)= \\ $$$$=\left(\mathrm{32}\:{mod}\:\mathrm{33}\right)^{\mathrm{19}} \:\centerdot\:\left(\mathrm{8}\:{mod}\:\mathrm{33}\right)= \\ $$$$=\left(\mathrm{32}\:{mod}\:\mathrm{33}\right)^{\mathrm{19}} \:\centerdot\:\left(\mathrm{8}\:{mod}\:\mathrm{33}\right)= \\ $$$$=\left[\left(−\mathrm{1}\right)^{\mathrm{19}} \:\centerdot\:\mathrm{8}\right]\:{mod}\:\mathrm{33}\:=\:\left(−\mathrm{8}\right){mod}\:\mathrm{33}= \\ $$$$=\left(\mathrm{33}−\mathrm{8}\right){mod}\:\mathrm{33}\:=\:\mathrm{25}\:{mod}\:\mathrm{33}\:=\:\mathrm{25} \\ $$

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 ⊛)

$$\left(\mathrm{1}\right)\:\mathcal{N}{ote}\:{that}\:{we}\:{have}\:\mathrm{2}^{{x}} = \\ $$$$\begin{cases}{\mathrm{1}\:\left({mod}\:\mathrm{3}\right)\:,\:{if}\:{x}\:{is}\:{even}\:}\\{\mathrm{2}\:\:\left({mod}\:\mathrm{3}\right),\:{if}\:{x}\:{is}\:{odd}\:}\end{cases} \\ $$$${Therefore}\:{we}\:{get}\:\underset{{n}\:=\:\mathrm{0}} {\overset{\mathrm{1000}} {\sum}}\lfloor\frac{\mathrm{2}^{{n}} }{\mathrm{3}}\rfloor\:=\:\mathrm{0} \\ $$$$+\:\underset{{n}\:=\:\mathrm{1}} {\overset{\mathrm{500}} {\sum}}\left(\lfloor\frac{\mathrm{2}^{\mathrm{2}{n}−\mathrm{1}} }{\mathrm{3}}\rfloor+\lfloor\frac{\mathrm{2}^{\mathrm{2}{n}} }{\mathrm{3}}\rfloor\right)\:=\:\underset{{n}\:=\:\mathrm{1}} {\overset{\mathrm{500}} {\sum}}\left(\frac{\mathrm{2}^{\mathrm{2}{n}−\mathrm{1}} −\mathrm{2}}{\mathrm{3}}+\frac{\mathrm{2}^{\mathrm{2}{n}} −\mathrm{1}}{\mathrm{3}}\right) \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{3}}\underset{{n}\:=\:\mathrm{1}} {\overset{\mathrm{500}} {\sum}}\left(\mathrm{2}^{\mathrm{2}{n}−\mathrm{1}} +\mathrm{2}^{\mathrm{2}{n}} −\mathrm{1}\right) \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{3}}\underset{{n}\:=\:\mathrm{1}} {\overset{\mathrm{1000}} {\sum}}\mathrm{2}^{{n}} −\mathrm{500}\: \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{2}^{\mathrm{1001}} −\mathrm{2}\right)−\mathrm{500}\:.\left({JS}\:\circledast\right) \\ $$

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