MCQ
The least remainder when ${17^{30}}$ is divided by $ 5$ is
- A$1$
- B$2$
- C$3$
- ✓$4$
==> ${({17^5})^6} = {(2)^6}$(mod $5$)
==> ${(17)^{30}} = 4$ (mod $5$)Hence required remainder $= 4.$
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$A:$$\cos \alpha + \cos \beta + \cos \gamma = 0$
$B$:$\sin \alpha + \sin \beta + \sin \gamma = 0$
If $\cos \left( {\alpha - \beta } \right) + \cos \left( {\beta - \gamma } \right) + \cos \left( {\gamma - \alpha } \right) = - \frac{3}{2}$ then