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remainder when 2^1040 is divided by 131
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- 2^1040 mod 131
Since 2 nd 131 are co-primes, we can use Euler theorem...
E[131] = 130
So 2^E[131] = 1
(2^130)^8 mod 131
1 mod 131
Ans : 1
- 10 years agoHelpfull: Yes(2) No(1)
- @SARASWATHY yrr whats E[131] = 130
i dont get this poin of u. not even in the previous ques of power - 10 years agoHelpfull: Yes(2) No(0)
- 2^1024 = 256^130
so its 256^130 mod 131
256=125(mod 131)
256^130=125^130(mod 131)
by Fermat’s Little Theorem
125^130=1(mod 131)
so remainder is 1 - 10 years agoHelpfull: Yes(1) No(0)
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