Question

Find the remainder when 1^2019 + 2^2019 + 3^2019 + ......2020^2019 is divided by 2019.​

Answer 1

Answer:

1

Step-by-step explanation:

the number is

1×1×1....(2019 times) + 2×2×2×...(2019 times)+.....+ 2020× 2020×2020...(2019 times)/2019

now there will be 1009 pairs of numbers less than 2019 who are divisible by 2019

1^2019 + 2018^2019 = (1+ 2018)(..) here the formula a^k + b^k = (a+b)(....) has been used

similarly we hv 2^2019 + 2017 ^2019 , ......, 1009^2019 + 1010^2019.

so now we know that all numbers up to 2018^2019 is divisible by 2019. & 2019^2019 is divisible clearly.

2020^1 = 1(mod 2019)

=> 2020^2019 = 1^2019(mod 2019)

=> 2020 ^2019 = 1(mod 2019)

so remaininder is 1

x=y( mod z )means that the x leaves same remainder as y leaves on division by z. This is in number theory