Question

Find the remainder when $2^{86}$ is divided by $7$

Answer 1

$\frac{2^{86}}{7}=\frac{2^{84}{.2}^2}{2^3-1}=\frac{{\left(2^3\right)}^{28}.4}{2^3-1}$
By remainder theorem, if $x=2^3$ and $f\left(x\right)=4x^{28}$
Required remainder $=$ $f\left(1\right)=4{\left(1\right)}^{28}=4$