Problem 10.4. Consider the sequence

$$
a_{n}=\cos (\underbrace{100 \ldots 0^{\circ}}_{n-1})
$$

For example, $a_{1}=\cos 1^{\circ}, a_{6}=\cos 100000^{\circ}$.

How many of the numbers $a_{1}, a_{2}, \ldots, a_{100}$ are positive?
Answer: 99.

Solution. Note that for an integer $x$ divisible by 40, the cosines of the angles $x^{\circ}$ and $10 x^{\circ}$ coincide, since the difference between these angles $9 x^{\circ}$ is divisible by $360^{\circ}$. Therefore, $\cos \left(\left(10^{k}\right)^{\circ}\right)=$ $\cos \left(\left(10^{k+1}\right)^{\circ}\right)$ for all $k \geqslant 3$.

It is clear that $\cos \left(1^{\circ}\right)>0, \cos \left(10^{\circ}\right)>0, \cos \left(100^{\circ}\right)<0$, and subsequently $0<\cos \left(1000^{\circ}\right)=\cos \left(10000^{\circ}\right)=\ldots=$ $\cos \left(\left(10^{99}\right)^{\circ}\right)$. Thus, among our numbers, exactly 99 are positive.