What is the remainder when 2²⁰¹⁹ is divided by 2019?
Answers
Answer:
604
Step-by-step explanation:
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Answer:
The answer is 8
Step-by-step explanation:
We can reform 2^2019/2019 as 2^2019/(3*673) where 3 and 673 are prime numbers and do not divide 2.
Thus these two can satisfy Fermat's little theorem, which says if a is prime and does not divides b then, b^(a-1) is congruent to 1 (mod a).
We will use this theorem along with Euclidean algorithm to solve it.
Using Fermat's little theorem, we can write
2^(673-1) is congruent to 1 ( mod 673)
=2^672 is congruent to 1( mod 673)
= 2^(672*3+3) is congruent to 2^3 ( mod 673)
=2^2019 is congruent to 8( mod 673)
Therefore the remainder when 2^2019 is divided by 673 is 8.
Similarly we will do with 3 :
2^(3-1) is congruent to 1 ( mod 3)
=2^2 is congruent to 1 ( mod 3)
= 2^(2*1009+1) is congruent to 2 ( mod 3)
= 2^ 2019 is congruent to 2 (mod 3)
Therefore the remainder in this case is 2.
Now using Euclidean algorithm we can write
673m+8=3n+2=2^2019
=> 673m=3n-6
=> m=3(n-2)/673
Let (n-2)/673 be k.
Using m =3k,
We can rewrite, 673m+8 as
673(3k)+8=2^2019
2019k +8=2^2019
Therefore when 2^2019 is divided by 2019 the remainder is 8.!!
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