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Find the remainder of (3^(9415))/(80)....

Find the remainder of `(3^(9415))/(80)`.

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Rem `3^9415/80=Rem(3^9412xx3^3)/80`
=Rem `(3^4)^2353xx3^3/80`
=Rem `(81^2353)xx3^3/80`=Rem`1xx27/80`=Rem27/80`
Thus the remainder is 27.
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