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

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

Grade:12th Pass

2 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
9 years ago
Ans:
Hello Student,
Please find answer to your question below

Let the remainder to be ‘R’
We have
\frac{2^{301}}{5}
\frac{2.2^{300}}{5}
\frac{2.(2^{3})^{100}}{5}
\frac{2.(8)^{100}}{5}
\frac{2.(10-2)^{100}}{5}
\frac{2.(5k-2)}{5}
2k-\frac{4}{5}
2k-1+1-\frac{4}{5}
(2k-1)+\frac{1}{5}
So remainder is 1.
dheeraj
40 Points
9 years ago
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)

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