$7^{{103}_{ }}$when divided by $25$leaves the remainder 

We have,

$\begin{array}{l} 7^{103}=7(49{)}^{51}\\ \Rightarrow 7^{103}=7(50-1{)}^{51}\\ \Rightarrow 7^{103}=7\left\{{ }^{51}{C}_0\left(50{)}^{51}{-}^{51}{C}_1\right(50{)}^{50}{+}^{51}{C}_2(50{)}^{49}\dots -1\right\}\\ \Rightarrow 7^{103}=7\left\{\left(50{)}^{51}{-}^{51}{C}_1\right(50{)}^{50}{+}^{51}{C}_2(50{)}^{49}\dots \right\}-7\\ \Rightarrow 7^{103}=7\left\{\left(50{)}^{51}{-}^{51}{C}_1\right(50{)}^{50}{+}^{51}{C}_2(50{)}^{49}\dots \right\}-7-18+18\\ \Rightarrow 7^{103}=7\left\{\left(50{)}^{51}{-}^{51}{C}_1\right(50{)}^{50}{+}^{51}{C}_2(50{)}^{49}\dots \right\}-25+18\end{array}$

$7^{{103}_{ }}$= Multiple of $25 + 18$

Hence, remainder = 18.