Problem (Number Theory):
For how many integers $k$ with $0<k<18$ is $\frac{5 \sin \left(10 k^{\circ}\right)-2}{\sin ^{2}\left(10 k^{\circ}\right)} \geq 2$ ?

Solution:
Let $\theta=10 k^{\circ}$.

The given inequalities become $0^{\circ}<\theta<180^{\circ}$ and $\frac{5 \sin \theta-2}{\sin ^{2} \theta} \geq 2$.

When $0^{\circ}<\theta<180^{\circ}, \sin \theta \neq 0$.

This means that we can can multiply both sides by $\sin ^{2} \theta>0$ and obtain the equivalent inequalities:

$$
\begin{aligned}
\frac{5 \sin \theta-2}{\sin ^{2} \theta} & \geq 2 \\
5 \sin \theta-2 & \geq 2 \sin ^{2} \theta \\
0 & \geq 2 \sin ^{2} \theta-5 \sin \theta+2 \\
0 & \geq(2 \sin \theta-1)(\sin \theta-2)
\end{aligned}
$$

Since $\sin \theta \leq 1$, then $\sin \theta-2 \leq-1<0$ for all $\theta$.

Therefore, $(2 \sin \theta-1)(\sin \theta-2) \leq 0$ exactly when $2 \sin \theta-1 \geq 0$.

Note that $2 \sin \theta-1 \geq 0$ exactly when $\sin \theta \geq \frac{1}{2}$.

Therefore, the original inequality is true exactly when $\frac{1}{2} \leq \sin \theta \leq 1$.

Note that $\sin 30^{\circ}=\sin 150^{\circ}=\frac{1}{2}$ and $0^{\circ}<\theta<180^{\circ}$.

When $\theta=0^{\circ}, \sin \theta=0$.

From $\theta=0^{\circ}$ to $\theta=30^{\circ}, \sin \theta$ increases from 0 to $\frac{1}{2}$.

From $\theta=30^{\circ}$ to $\theta=150^{\circ}, \sin \theta$ increases from $\frac{1}{2}$ to 1 and then decreases to $\frac{1}{2}$.

From $\theta=150^{\circ}$ to $\theta=180^{\circ}, \sin \theta$ decreases from $\frac{1}{2}$ to 0 .

Therefore, the original inequality is true exactly when $30^{\circ} \leq \theta \leq 150^{\circ}$ which is equivalent to $30^{\circ} \leq 10 k^{\circ} \leq 150^{\circ}$ and to $3 \leq k \leq 15$.

The integers $k$ in this range are $k=3,4,5,6, \ldots, 12,13,14,15$, of which there are 13 .

Final answer: 13