Let's look at FERMAT's LITTLE THEOREM before we proceed to answer to this....
Let P be any prime number and N be a number which is not divisible by P.
Then the remainder obtained when N(P-1) divided by P is 1.
So applying this rule here ( since 101 is prime) and 104 is not divisible by 101,
we can write the given number as
(104100)3 x (1043)
and (104100)3 leave a remainder 1 according to Fermat's rule. (i.e 104100 gives rem 1)
So now the problem finally becomes 1043 which leaves remainder 27 when divided by 101.
I hope I am clear.
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