Remainder Problem - 2

Method-1 : \[\begin{gathered} R\left( {\frac{{{3^{98}}}}{5}} \right) = ? \hfill \\ = R\left( {\frac{{{3^{2 \times 49}}}}{5}} \right) \hfill \\ = R\left( {\frac{{{{\left( {{3^2}} \right)}^{49}}}}{5}} \right) \hfill \\ = R\left( {\frac{{{{\left( { - 1} \right)}^{49}}}}{5}} \right) \hfill \\ = R\left( {\frac{{ - 1}}{5}} \right) \hfill \\ = R\left( {\frac{{5 - 1}}{5}} \right) \hfill \\ = 4 \hfill \\ \therefore R\left( {\frac{{{3^{98}}}}{5}} \right) = 4 \hfill \\ \end{gathered} \] Method- 2: We will be using Fermat’s Little Theorem to solve this problem: \[\begin{gathered} {a^{p - 1}} \equiv \left( 1 \right)\,\bmod \,\left( p \right),p:prime \hfill \\ {3^4} \equiv \left( 1 \right)\,\bmod \,\left( 5 \right) \hfill \\ \Rightarrow {\left( {{3^4}} \right)^{24}} \equiv {\left( 1 \right)^{24}}\,\bmod \,\left( 5 \right) \hfill \\ \Rightarrow \left( {{3^{96}}} \right) \equiv \left( 1 \right)\,\bmod \,\left( 5 \right) \hfill \\ \Rightarrow \left( {{3^{96}} \times {3^2}} \right) \equiv \left( {1 \times {3^2}} \right)\,\bmod \,\left( 5 \right) \hfill \\ \Rightarrow {3^{98}} \equiv \left( 9 \right)\,\bmod \,\left( 5 \right) \hfill \\ \Rightarrow {3^{98}} \equiv \left( 4 \right)\,\bmod \,\left( 5 \right) \hfill \\ \therefore R\left( {\frac{{{3^{98}}}}{5}} \right) = 4 \hfill \\ \end{gathered} \]