1. How many positive numbers are there among the 2014 terms of the sequence: $\sin 1^{\circ}, \sin 10^{\circ}$, $\sin 100^{\circ}, \sin 1000^{\circ}, \ldots ?$
Solution. For $n>3$ we have

$10^{n}-1000=10^{3}\left(10^{n-3}-1\right)=25 \cdot 40 \cdot\left(10^{n-3}-1\right)$.

Since $10^{n-3}-1$ is divisible by 9, it follows that $10^{n}-1000$ is divisible by 360. Therefore, all terms of the sequence, starting from the fourth, coincide with $\sin 1000^{\circ}=\sin \left(3 \cdot 360^{\circ}-80^{\circ}\right)=\sin \left(-80^{\circ}\right)<0$. Thus, the positive terms in the sequence are $\sin 1^{\circ}, \sin 10^{\circ}, \sin 100^{\circ}$.