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


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

$(1) E v a l u a t e t h e s u m ⌊ \frac{2^{0}}{3} ⌋ + ⌊ \frac{2^{1}}{3} ⌋ + ⌊ \frac{2^{2}}{3} ⌋ + . . . + ⌊ \frac{2^{1000}}{3} ⌋$ $(2) f i n d 2^{98} (m o d 33)$
	Answered by JDamian last updated on 19/Jul/20
	$(2) 2^(98) mod 33=[(2^5 )^(19) 2^3 ]mod 33= ={[(2^5 )^(19) ]mod 33} ∙ (8 mod 33)= ={[(32)^(19) ]mod 33} ∙ (8 mod 33)= =(32 mod 33)^(19) ∙ (8 mod 33)= =(32 mod 33)^(19) ∙ (8 mod 33)= =[(−1)^(19) ∙ 8] mod 33 = (−8)mod 33= =(33−8)mod 33 = 25 mod 33 = 25$
	$(2) 2^{98} m o d 33 = [{(2^{5})}^{19} 2^{3}] m o d 33 =$ $= {[{(2^{5})}^{19}] m o d 33} \cdot (8 m o d 33) =$ $= {[{(32)}^{19}] m o d 33} \cdot (8 m o d 33) =$ $= {(32 m o d 33)}^{19} \cdot (8 m o d 33) =$ $= {(32 m o d 33)}^{19} \cdot (8 m o d 33) =$ $= [{(- 1)}^{19} \cdot 8] m o d 33 = (- 8) m o d 33 =$ $= (33 - 8) m o d 33 = 25 m o d 33 = 25$
	Answered by john santu last updated on 19/Jul/20
	$(1) Note that we have 2^x = { ((1 (mod 3) , if x is even )),((2 (mod 3), if x is odd )) :} Therefore we get Σ_(n = 0) ^(1000) ⌊(2^n /3)⌋ = 0 + Σ_(n = 1) ^(500) (⌊(2^(2n−1) /3)⌋+⌊(2^(2n) /3)⌋) = Σ_(n = 1) ^(500) (((2^(2n−1) −2)/3)+((2^(2n) −1)/3)) = (1/3)Σ_(n = 1) ^(500) (2^(2n−1) +2^(2n) −1) = (1/3)Σ_(n = 1) ^(1000) 2^n −500 = (1/3)(2^(1001) −2)−500 .(JS ⊛)$
	$(1) N o t e t h a t w e h a v e 2^{x} =$ ${\begin{cases} 1 (m o d 3), i f x i s e v e n \\ 2 (m o d 3), i f x i s o d d \end{cases}$ $T h e r e f o r e w e g e t \underset{n = 0}{\sum^{1000}} ⌊ \frac{2^{n}}{3} ⌋ = 0$ $+ \underset{n = 1}{\sum^{500}} (⌊ \frac{2^{2 n - 1}}{3} ⌋ + ⌊ \frac{2^{2 n}}{3} ⌋) = \underset{n = 1}{\sum^{500}} (\frac{2^{2 n - 1} - 2}{3} + \frac{2^{2 n} - 1}{3})$ $= \frac{1}{3} \underset{n = 1}{\sum^{500}} (2^{2 n - 1} + 2^{2 n} - 1)$ $= \frac{1}{3} \underset{n = 1}{\sum^{1000}} 2^{n} - 500$ $= \frac{1}{3} (2^{1001} - 2) - 500 . (J S ⊛)$
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