Math Question and Answers


Question and Answers Forum

All Questions Topic List
Arithmetic Questions
Previous in All Question Next in All Question
Previous in Arithmetic Next in Arithmetic
Question Number 35150 by Victor31926 last updated on 16/May/18

	Answered by Rasheed.Sindhi last updated on 16/May/18

	$∙ Any power of even number is even$ $and even number gives remainder 0$ $on dividing by 2$ $∙ Any power of odd number is odd$ $and odd number gives remainder 1$ $on dividing by 2$ $\overset{∙}{∙ ∙}$ $1^{2018} \equiv 1 (\mod 2)$ $2^{2018} \equiv 0 (\mod 2)$ $3^{2018} \equiv 1 (\mod 2)$ $4^{2018} \equiv 0 (\mod 2)$ $5^{2018} \equiv 1 (\mod 2)$ $6^{2018} \equiv 0 (\mod 2)$ $7^{2018} \equiv 1 (\mod 2)$ $8^{2018} \equiv 0 (\mod 2)$ $Adding all above congruences$ $1^{2018} + 2^{2018} + . . . + 8^{2018}$ $\equiv 1 + 0 + 1 + 0 + 1 + 0 + 1 + 0 (\mod 2)$ $\equiv 4 \equiv 0 (\mod 2)$ $\overset{∙}{∙ ∙} The remainder is 0$
	Answered by tanmay.chaudhury50@gmail.com last updated on 16/May/18

	$l e t s o l v e i t a n a l y t i c a l l y . . .$ $s e g r e g a t i n g t h e t e r m s$ $(1^{2018} + 3^{2018} + 5^{2018} + 7^{2018}) +$ $(2^{2018} + 4^{2018} + 6^{2018} + 8^{2018})$ $t h e e v e n t e r m s s e g r e g a t e d i n b r a c k e t s a r e$ $d i v i s b l e b y 2$ $n o w$ $7^{2018} = {(8 - 1)}^{2018}$ $= 8^{2018} - C_{1}^{2018} . 8^{2017} + C_{2}^{2018} 8^{2016} . . . . + {(- 1)}^{2018} s o$ $w h e n d e v i d e d b y 2 R e m a i n d e r i s . . . {(- 1)}^{2018} = 1$ $5^{2018} = {(6 - 1)}^{2018} = 6^{2018} - C_{1}^{2018} 6^{2017} + . . . + {(- 1)}^{2018}$ $s o R = {(- 1)}^{2018} = 1$ $3^{2018} = {(4 - 1)}^{2018} = 4^{2018} - C_{1}^{2018} . 4^{2017} + . . . + {(- 1)}^{2018}$ $s o R = {(- 1)}^{2018} = 1$ ${(1)}^{2018} = 1$ $s o s u m o f r e m a i n d e r s = 1 + 1 + 1 + 1 = 4$ $d i v i s i b l e b y 2 s o R e m a i n d e r = 0$ $s o w h e n 1^{2018} + 2^{2018} + . . . + 8^{2018} i s d e v i d e d b y 2$ $r e m a i n d e r i s 0$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum