Question

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

Answer 1

Answer: 01

$[3^{400}=81^{100}=(1+80{)}^{100}$
$={}^{100}{C}_0+{}^{100}{C}_{1}80+\dots$ $+{}^{100}{C}_{100}80^{100}$
$\Rightarrow$ Last two digits are $01$