If 1^2019 +2^2019 +3^2019 ..... +2020^2019 is divided by 2019 , what will be the remainder. Please solve using Congruence and Modulo
Answers
Answer:
1
Fermat’s little theorem: a^(p-1), when divided by p the remainder is 1, where p is prime and p does not divide a.
when 2020^2019 divided by 2019, the remainder is 1.
When 2019^2019 divided by 2019, the remainder is 0.
When 2018^2019 divided by 2019, the remainder is -1.
When 2017^2018*2017 divided by 2019, the remainder is -2 and so on.
The sum equals 1. Therefore, the answer is 1.
Step-by-step explanation:
a^n+b^n=(a+b)(…..) for odd ‘n’
(a+b) is a factor of a^n+b^n: with b=-a,a^n+b^n vanishes for odd n
We group as follows
2019=(1+2018) is a factor of 1^2019+2018^2019
2019=(2+2017) is a factor of 2^2019+2017^2019
2019=(3+2016) is a factor of 3^2019+2016^2019
………………………………………………………………..
[From 1^2019 to 2019^2018 there are 2018 terms , an even number of terms: there are two middle terms]
2019 is a factor of 2019^2019
Now 2020^2019=(2019+1)^2019
From binomial expansion all terms with the exception of unity are exactly divisible by 2019
Therefore,remainder=1
With a bit of variation:
The given sum that is 1^2019+2^2019+…….2019^2019+2020^2019 is exactly divisible by 2020is exactly divisible by 2020
1^2019+2019^2019 contains 2020=1+2019 a s a factor
2^2019+2018^2019 contains 2020=2+2018 as a factor
1010^2019 is a solitary term if we consider first terms from 1^2019 to 2019^2019. Nevertheless it is divisible by 2020[1010^2019=1010*1010*1010^2017 is exactly divisible by 2020]
………….
2020^2019 is exactly divisible by 2020
Therefore the sum:1^2019+2^2019+…….2019^2019+2020^2019 is exactly divisible by 2020
Step-by-step explanation:
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