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Question-35150

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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 ••^(•) 1^(2018) ≡1(mod 2) 2^(2018) ≡0(mod 2) 3^(2018) ≡1(mod 2) 4^(2018) ≡0(mod 2) 5^(2018) ≡1(mod 2) 6^(2018) ≡0(mod 2) 7^(2018) ≡1(mod 2) 8^(2018) ≡0(mod 2) Adding all above congruences 1^(2018) +2^(2018) +...+8^(2018) ≡1+0+1+0+1+0+1+0(mod 2) ≡4≡0(mod 2) ••^(•) The remainder is 0
$$\bullet\mathrm{Any}\:\mathrm{power}\:\mathrm{of}\:\mathrm{even}\:\mathrm{number}\:\mathrm{is}\:\mathrm{even} \\ $$$$\:\:\:\:\:\mathrm{and}\:\mathrm{even}\:\mathrm{number}\:\mathrm{gives}\:\mathrm{remainder}\:\mathrm{0} \\ $$$$\:\:\:\:\:\mathrm{on}\:\mathrm{dividing}\:\mathrm{by}\:\mathrm{2} \\ $$$$\bullet\mathrm{Any}\:\mathrm{power}\:\mathrm{of}\:\mathrm{odd}\:\mathrm{number}\:\mathrm{is}\:\mathrm{odd} \\ $$$$\:\:\:\:\:\mathrm{and}\:\mathrm{odd}\:\mathrm{number}\:\mathrm{gives}\:\mathrm{remainder}\:\mathrm{1} \\ $$$$\:\:\:\:\:\mathrm{on}\:\mathrm{dividing}\:\mathrm{by}\:\mathrm{2} \\ $$$$\:\overset{\bullet} {\bullet\bullet} \\ $$$$\mathrm{1}^{\mathrm{2018}} \equiv\mathrm{1}\left(\mathrm{mod}\:\mathrm{2}\right) \\ $$$$\mathrm{2}^{\mathrm{2018}} \equiv\mathrm{0}\left(\mathrm{mod}\:\mathrm{2}\right) \\ $$$$\mathrm{3}^{\mathrm{2018}} \equiv\mathrm{1}\left(\mathrm{mod}\:\mathrm{2}\right) \\ $$$$\mathrm{4}^{\mathrm{2018}} \equiv\mathrm{0}\left(\mathrm{mod}\:\mathrm{2}\right) \\ $$$$\mathrm{5}^{\mathrm{2018}} \equiv\mathrm{1}\left(\mathrm{mod}\:\mathrm{2}\right) \\ $$$$\mathrm{6}^{\mathrm{2018}} \equiv\mathrm{0}\left(\mathrm{mod}\:\mathrm{2}\right) \\ $$$$\mathrm{7}^{\mathrm{2018}} \equiv\mathrm{1}\left(\mathrm{mod}\:\mathrm{2}\right) \\ $$$$\mathrm{8}^{\mathrm{2018}} \equiv\mathrm{0}\left(\mathrm{mod}\:\mathrm{2}\right) \\ $$$$\mathrm{Adding}\:\mathrm{all}\:\mathrm{above}\:\mathrm{congruences} \\ $$$$\mathrm{1}^{\mathrm{2018}} +\mathrm{2}^{\mathrm{2018}} +…+\mathrm{8}^{\mathrm{2018}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\equiv\mathrm{1}+\mathrm{0}+\mathrm{1}+\mathrm{0}+\mathrm{1}+\mathrm{0}+\mathrm{1}+\mathrm{0}\left(\mathrm{mod}\:\mathrm{2}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\equiv\mathrm{4}\equiv\mathrm{0}\left(\mathrm{mod}\:\mathrm{2}\right) \\ $$$$\overset{\bullet} {\bullet\bullet}\mathrm{The}\:\mathrm{remainder}\:\mathrm{is}\:\mathrm{0} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 16/May/18
let solve it analytically... segregating the terms (1^(2018) +3^(2018) +5^(2018) +7^(2018) )+ (2^(2018) +4^(2018) +6^(2018) +8^(2018) ) the even terms segregated in brackets are divisble by 2 now 7^(2018) =(8−1)^(2018) =8^(2018) −C_1 ^(2018) .8^(2017) +C_2 ^(2018) 8^(2016) ....+(−1)^(2018) so when devided by 2 Remainder is...(−1)^(2018) =1 5^(2018) =(6−1)^(2018) =6^(2018) −C_1 ^(2018) 6^(2017) +...+(−1)^(2018) so R=(−1)^(2018) =1 3^(2018) =(4−1)^(2018) =4^(2018) −C_1 ^(2018) .4^(2017) +...+(−1)^(2018) so R=(−1)^(2018) =1 (1)^(2018) =1 so sum of remainders=1+1+1+1=4 divisible by 2 so Remainder=0 so when 1^(2018) +2^(2018) +...+8^(2018) is devided by 2 remainder is 0
$${let}\:{solve}\:{it}\:{analytically}… \\ $$$${segregating}\:{the}\:{terms} \\ $$$$\left(\mathrm{1}^{\mathrm{2018}} +\mathrm{3}^{\mathrm{2018}} +\mathrm{5}^{\mathrm{2018}} +\mathrm{7}^{\mathrm{2018}} \right)+ \\ $$$$\left(\mathrm{2}^{\mathrm{2018}} +\mathrm{4}^{\mathrm{2018}} +\mathrm{6}^{\mathrm{2018}} +\mathrm{8}^{\mathrm{2018}} \right) \\ $$$${the}\:{even}\:{terms}\:{segregated}\:{in}\:{brackets}\:{are} \\ $$$${divisble}\:{by}\:\mathrm{2} \\ $$$${now}\: \\ $$$$\mathrm{7}^{\mathrm{2018}} =\left(\mathrm{8}−\mathrm{1}\right)^{\mathrm{2018}} \\ $$$$=\mathrm{8}^{\mathrm{2018}} −{C}_{\mathrm{1}} ^{\mathrm{2018}} .\mathrm{8}^{\mathrm{2017}} +{C}_{\mathrm{2}} ^{\mathrm{2018}} \mathrm{8}^{\mathrm{2016}} ….+\left(−\mathrm{1}\right)^{\mathrm{2018}} \:{so} \\ $$$$\:{when}\:{devided}\:{by}\:\mathrm{2}\:{Remainder}\:{is}…\left(−\mathrm{1}\right)^{\mathrm{2018}} =\mathrm{1} \\ $$$$\mathrm{5}^{\mathrm{2018}} =\left(\mathrm{6}−\mathrm{1}\right)^{\mathrm{2018}} =\mathrm{6}^{\mathrm{2018}} −{C}_{\mathrm{1}} ^{\mathrm{2018}} \mathrm{6}^{\mathrm{2017}} +…+\left(−\mathrm{1}\right)^{\mathrm{2018}} \\ $$$${so}\:{R}=\left(−\mathrm{1}\right)^{\mathrm{2018}} =\mathrm{1} \\ $$$$\mathrm{3}^{\mathrm{2018}} =\left(\mathrm{4}−\mathrm{1}\right)^{\mathrm{2018}} =\mathrm{4}^{\mathrm{2018}} −{C}_{\mathrm{1}} ^{\mathrm{2018}} .\mathrm{4}^{\mathrm{2017}} +…+\left(−\mathrm{1}\right)^{\mathrm{2018}} \\ $$$${so}\:{R}=\left(−\mathrm{1}\right)^{\mathrm{2018}} =\mathrm{1} \\ $$$$\:\left(\mathrm{1}\right)^{\mathrm{2018}} =\mathrm{1} \\ $$$${so}\:{sum}\:{of}\:{remainders}=\mathrm{1}+\mathrm{1}+\mathrm{1}+\mathrm{1}=\mathrm{4} \\ $$$${divisible}\:{by}\:\mathrm{2}\:\:{so}\:{Remainder}=\mathrm{0} \\ $$$${so}\:{when}\:\mathrm{1}^{\mathrm{2018}} +\mathrm{2}^{\mathrm{2018}} +…+\mathrm{8}^{\mathrm{2018}} \:{is}\:{devided}\:{by}\:\mathrm{2} \\ $$$${remainder}\:{is}\:\mathrm{0} \\ $$

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