Question

Find the last two digit of $3^{400}?$

Answer 1

We have, $3^{400}={\left(3^2\right)}^{200}=9^{200}={(10-1)}^{200}$

$={\left(10\right)}^{200}{-}^{200}{C}_1{\left(10\right)}^{199}{+}^{200}{C}_2{\left(10\right)}^{198}{-}^{200}{C}_3{\left(10\right)}^{197}+...{+}^{200}{C}_{198}{\left(10\right)}^2{-}^{200}{C}_{199}\left(10\right)+1$

$=100\mu {-}^{200}{C}_{199}\left(10\right)+1$ where $\mu \in I$

$=100\mu {-}^{200}{C}_1\left(10\right)+1$

$=100\mu -2000+1$

$=100(\mu -20)+1$

$=100p+1$ where $p$ is an integer.

Hence, last two digits of $3^{400}$ is $00+1=01$