Question

If $(2021{)}^{3762}$ is divided by $17$, then the remainder is

Answer 1

To find remainder when $\mathbf{(}\mathbf{2021}{\mathbf{)}}^{\mathbf{3762}}$ is divided by $\mathbf{17}$:

$\begin{array}{rl}(2021{)}^{3762}& =(2023-2{)}^{3762}\\ & =17\lambda +2^{3762}\end{array}$

where $\lambda$ is a constant.

$\begin{array}{rl}17\lambda +2^{3762}& =17\lambda +2^2(2^8{)}^{470}\\ & =17\lambda +2^2(2^8{)}^{470}\\ & =17\lambda +4(17\mu +1{)}^{470}\\ & =17k+4\end{array}$

where $\mu$ and $k$ are constants.

Hence, the remainder, when $\mathbf{(}\mathbf{2021}{\mathbf{)}}^{\mathbf{3762}}$ is divided by $\mathbf{17}$ is $\mathbf{4}$.