Question

The remainder obtained when $27^{40}$ is divided by $12$ is

Answer 1

$\left(27\right)^{40}=\left(3^{3}\right)^{40}=3^{120}$
$3^{119}=\left(4 - 1\right)^{119}$
$=\_{}^{119}C_{0}4^{119}-\_{}^{119}C_{1}4^{118}+\_{}^{119}C_{2}4^{117}-.\ldots \ldots .+\_{}^{119}C_{118}4-\_{}^{119}C_{119}$
$=4k-1$
$\Rightarrow 3^{120}=3\left(4 k - 1\right)=12k-3$
$=12\left(k - 1\right)+9$
So, the required remainder is $9$ .