Math, asked by kishlay0123, 1 year ago

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

Answered by sivaprasath
39

Answer:

1

Step-by-step explanation:

Given :

To find the remainder,

when,

1^{2019} + 2^{2019} + 3^{2019} + .... + 2020^{2019} is divided by 2019,.

Solution :

We know that,

for any two numbers with same odd powers, their sum is divisible by them,.

i.e.,

Let a & b be two integers,.

then,. iff n is odd , a^n + b^n is divisible by a + b,.

Let a + b = 2019 , n = 2019,.

1 + 2018 = 2019,1^{2019} + 2018^{2019} is divisible by 2019,.

2 + 2017 = 2019,2^{2019} + 2017^{2019} is divisible by 2019,.

Then, by dividing the numbers into suitable pairs,

[1^{2019} + 2^{2019} + 3^{2019} + .... + 2018^{2019}] + 2019^{2019} + 2020^{2020}

[(1^{2019} + 2018^{2019}) + (2^{2019} + 2017^{2019}) + (3^{2019} + 2016^{2019})+ ... (1009^{2019}+1010^{2019}) + 2019^{2019}+2020^{2019})

[(1^{2019} + 2018^{2019}) + (2^{2019} + 2017^{2019}) + (3^{2019} + 2016^{2019})+ ... (1009^{2019}+1010^{2019}) is divisible by 2019,.

also 2019^{2019} is divisible by 2019,.

⇒ Hence,. The last 2020²⁰¹⁹ can be written as,.

(by adding and subtracting (-1)²⁰¹⁹)

[(1^{2019} + 2018^{2019}) + (2^{2019} + 2017^{2019}) + (3^{2019} + 2016^{2019})+ ... (1009^{2019}+1010^{2019}) +[2019^{2019} + 0^{2019}) +[2020^{2019} + (-1)^{2019} ] - (-1)^{2019}]

As,  (-1)²⁰¹⁹ = -1

[(1^{2019} + 2018^{2019}) + (2^{2019} + 2017^{2019}) + (3^{2019} + 2016^{2019})+ ... (1009^{2019}+1010^{2019}) +[2019^{2019} + 0^{2019}) +[2020^{2019} + (-1)^{2019}] is divisible by 2019,.

(-1)²⁰¹⁹ = -1 ,

There - (-1)²⁰¹⁹ = -(-1)= 1 ,. is remaining (extra)

hence, the remainder is 1,.


sivaprasath: nice Question, very tricky,.
Answered by p9999
6

Answer:

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