Question:
If $(2021)^{3762}$ is divided by 17 , then the remainder is
Solution:
$(2023-2)^{3762}=2023 \mathrm{k}_{1}+2^{3762}$
$=17 \mathrm{k}_{2}+2^{3762}($ as $2023=17 \times 17 \times 9)$
$=17 \mathrm{k}_{2}+4 \times 16^{940}$
$=17 \mathrm{k}_{2}+4 \times(17-1)^{940}$
$=17 \mathrm{k}_{2}+4\left(17 \mathrm{k}_{3}+1\right)$
$=17 \mathrm{k}+4 \Rightarrow$ remainder $=4$