The remainder when $9^{103}$ is divided by $25$ is equal to 

We have,

$9^{103}$

$\begin{array}{l}=9\times {\left(81\right)}^{51}\\ =9(80+1{)}^{51}\\ =\begin{array}{c} \\ 9\left[\left(80{)}^{51}{+}^{51}{C}_1\right(80{)}^{50}{+}^{51}{C}_2(80{)}^{49}+\dots {+}^{51}{C}_{50}(80)+1\right]\end{array}\\ =9(25k+(51\times 80+1\left)\right\}\end{array}$

$\begin{array}{l}=25\lambda +36729\\ =25\lambda +25\times 1469+4\\ =25(\lambda +1469)+4\end{array}$

Hence, the remainder is $4$