1-Evaluate-the-sum-2-0-3-2-1-3-2-2-3-2-1000-3-2-find-2-98-mod-33-

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} ⌋ + \dots + ⌊ \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} 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) 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 ⊛)