Question
The remainder, when $3^{2003}$ is divided by $28$, is
✓
Answer
c
$\frac{\left(3^{3}\right)^{667} \cdot 3^{2}}{28}=\frac{(28-1)^{667} \cdot 3^{2}}{28}=-9+28=19$
$\frac{\left(3^{3}\right)^{667} \cdot 3^{2}}{28}=\frac{(28-1)^{667} \cdot 3^{2}}{28}=-9+28=19$
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