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

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(4) : $(2021)^{3762}=(2023-2)^{3762}$\ $=17 a+2^{3762}$, where $17 a$ is multiple of 17 .\ Now, $2^{3762}=4 \cdot 16^{940}=4(17-1)^{940}$\ $=4(1-17)^{940}=17 b+4$\ $	herefore \quad$ Required remainder $=4$\