The remainder, when 3^ 2003 is divided by 28, is - Vidyadip

Question

The remainder, when $3^{2003}$ is divided by $28$, is

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Answer

c
$\frac{\left(3^{3}\right)^{667} \cdot 3^{2}}{28}=\frac{(28-1)^{667} \cdot 3^{2}}{28}=-9+28=19$

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