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What will be the remainder when 1^2017+2^2017+3^2017+4^2017..................2018^2017/2017 |
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Answer» This one is good... 2,019 IS NOT PRIME... I NEED Euler’s help... φ(2,019)=2,019×(2/3)×(672/673)=1,344【2019=3×673】 ∴Euler said : a^1344 ≡ 1 mod 2019 Now note that 2017=1344+673 I need a PATTERN..There are 2,017 terms... Working from the front... ALL in mod...2019 1^2017≡ 1 2^2017≡2^673 3^2017≡3^673 4^2017≡4^673 ......... 99^2017≡99^673 1008^2017≡1008^673 .............… Working from the back... 2017^2017≡2017^673≡(2019–2)^673≡-(2^673) 2016^2017≡2016^673≡-(3^673) 2015^2017≡2015^673≡-(4^673).... ............... 1920^2017≡(2019–99)^673≡ -(99)^673 1008^2017≡(2019–1011)^673≡-1011^673 1018^2017≡(2019–1001)^2017≡-(1001^673) 1019^2017≡(2019–1000)^673≡-(1000^673). OBSERVATION : 2nd cancell off with 2017th term 3rd cancell off with 2016th term .........and so on ∴The only uncanclled term is the FIRST TERM. ∴THE GIVEN series ≡ 1 mod 2019 |
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