Question
The remainder when $428^{2024}$ is divided by $21$ is $.......$

Answer

$(428)^{2024}=(420+8)^{2024}$
$=(21 \times 20+8)^{2024}$
$=21 m+8^{2024}$
Now $8^{2024}=\left(8^2\right)^{1012}$
$=(64)^{1012}$
$=(63+1)^{1012}$
$=(21 \times 3+1)^{1012}$
$=2 \ln +1$
$\Rightarrow \text { Remainder is } 1$

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