what will be the remainder when 1^2017+2^2017+3^2017+4^2017..................2018^2017/2017
Answers
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