Question

Write last two digits of the number $3^{400}$

Answer 1

$3^{400}=(9{)}^{200}$

$=(10-1{)}^{200}$

${=}^{200}{C}_0\left(10{)}^{200}{+}^{200}{C}_1\right(10{)}^{199}(-1{)}^1+...{.}^{200}{C}_{198}(10{)}^2(-1{)}^{198}{+}^{200}{C}_{199}(10{)}^1(-1{)}^{199}{+}^{200}{C}_{200}(-1{)}^{200}$

$100\left[\right(10{)}^{198}{+}^{200}{C}_1\left(10{)}^{197}\right(-1{)}^1+....{+}^{200}{C}_{198}(-1{)}^{198}]+200\left(10{)}^1\right(-1{)}^{199}{+}^{200}{C}_{200}(-1{)}^{200}$

$=100\left[\right(10{)}^{198}{-}^{200}{C}_1(10{)}^{197}+....{+}^{200}{C}_{198}-2(10\left)\right]+1$

=100 (a natural number)+1

Hence, last two digits of the number $3^{400}$ is 01