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Answers
Answer:
Answer:
1
Step-by-step explanation:
Given :
To find the remainder,
when,
⇒ 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 , is divisible by a + b,.
Let a + b = 2019 , n = 2019,.
⇒ ⇒ is divisible by 2019,.
⇒ ⇒ is divisible by 2019,.
Then, by dividing the numbers into suitable pairs,
⇒
⇒ is divisible by 2019,.
also is divisible by 2019,.
⇒ Hence,. The last 2020²⁰¹⁹ can be written as,.
(by adding and subtracting (-1)²⁰¹⁹)
⇒
As, (-1)²⁰¹⁹ = -1
⇒ is divisible by 2019,.
(-1)²⁰¹⁹ = -1 ,
There - (-1)²⁰¹⁹ = -(-1)= 1 ,. is remaining (extra)
hence, the remainder is 1,.
Answer:
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’.
2019 = (1+2018) is a factor of 1^2019 + 2018^2019
2019 = (2+2017) is a factor of 2^2019 + 2017^2019
2019 = (3+2017) is a factor of 3^2019 + 2016^2019
From 1^2019 to 2019^2018, there are 2018 terms, am even number of terms and thee 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.