Question

fimd the remainder when 1²⁰¹⁹ + 2²⁰¹⁹ ...... 2020²⁰¹⁹ is divided by 2019​

Answer 1

Given: The expression 1²⁰¹⁹ + 2²⁰¹⁹ ...... 2020²⁰¹⁹

To find: The remainder when expression is divided by 2019​.

Solution:

• Now lets expand the given expression a little, we get:

              1^2019 + 2^2019 + 3^2019  +......+ 2016^2019 + 2017^2019 + 2018^2019   + 2019^2019  + 2020^2019

We can write it as:

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

• Now we know that, a^n + b^n  is divisible by ( a + b ), by the condition that n is odd.
• So in that case the given expression is divisible by 2019.

• So,

              2020^2019   =  ( 1 + 2019 )^2019

• By applying combination we get:

              1+  ( 2019  C  1 ) x 2019+   (2019  C  2)  x 2019^2  +....+ ( 2019 C 2019  ) x 2019^2019

• Taking 2019 common, we get:

              2019 x (2019  C  1  +   2019  C  2 x 2019+....+ 2019 C 2018 ) + 1

• So, 2020^ 2019  when divided by 2019 leaves the remainder as 1.

Answer:

                  So the remainder comes out to be 1.