Question

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

• A. $5$
• B. $6$
• C. $4$
• D. None of these

Answer 1

Answer: (C)

$4$

$9^{103}=9\times 9^{102}=9.(81{)}^{51}=9.(80+1{)}^{51}$

On expanding $(80+1{)}^{51}$ , We get

$9\left(\underset{divisibleby25}{\underset{}{{}^{51}{C}_0\left(80{)}^{51}{+}^{51}{C}_1\right(80{)}^{50}+\dots }}{+}^{51}{C}_{50}\right(80{)}^1+1)$

$25\lambda +(51\times 80\times 9)+9=25{\lambda }_1+36729=25{\lambda }_2+4\left(36729givesaremainderof4whendividedby25\right)$

Remainder $=4$