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

Binomial Theorem

Solution:

$27^{40}=3^{120} \Rightarrow 3^{119}=(4-1)^{119}$
$={ }^{119} C _{0} 4^{119}-{ }^{119} C _{1} 4^{118}$
$+119 C _{2} 4^{117}-{ }^{119} C _{3} 4^{116}+\ldots .+(-1)$
$\therefore 3^{119}=4 k-1 $
$ \therefore 3^{120}=12\, k-3=12(k-1)+9$
$\therefore $ The required remainder is $9$