MCQ
The remainder when $(2021)^{203}$ is divided by $7$ is
- A$1$
- B$2$
- ✓$5$
- D$6$
$={ }^{2023} C_{0}(7 A )^{2023}-\ldots{ }^{2023} C _{2023} 2^{2023}$
$=7 t -2^{2023}$
$\therefore-2^{2023}=-2 \times 2^{2022}$
$=-2 \times\left(2^{3}\right)^{674}$
$=-2(1+7 \mu)^{674}$
$=-(7 \alpha+2)$
$\Rightarrow \text { remainder }=-2 \text { or }+5$
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