Question

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

Answer 1

Given : $(2021{)}^{3762}$
We know that
$2023=17\times 119$
Now,
$(2021{)}^{3762}=(2023-2{)}^{3762}=(-2+2023{)}^{3762}=\sum _{r=0}^{3762}{}^{3762}{C}_r(-2{)}^{3762-r}(2023{)}^r={}^{3762}{C}_0(-2{)}^{3762}(2023{)}^0+17\lambda ,\lambda \in I=17\lambda +2^2(2^4{)}^{940}=17\lambda +4(17-1{)}^{940}=17\lambda +4(17\mu +1)=17k+4,k\in I$

Hence, the remainder is $4.$