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Find the remainder of (3^(9415))/(80). |
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Answer» Solution :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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