what is the remainder when 2 to the power of 301 is divided by 5

wkt (2,1)common factor is 1

so, we take 2^301-1congruent to 1(mod301)

=2^300 congruent to 1(mod301)            (let it be eqn 1)

wkt   2 is congruent to 1 (mod301)           (let it be eqn 2)

now multiply both the eqn’s and we get

  2^301 is congruent to 1(mod301)                                                             [note  here any integer before (mod n) is remainder of given number]

so,we get 1 as remainder of the given equation(eqn)