[Number Theory]
Problem: Let $S$ be the set of prime factors of the numbers you receive from positions 7 and 9 , and let $p$ and $q$ be the two least distinct elements of $S$, with $p<q$. Hexagon HEXAGO is inscribed in circle $\omega$, and every angle of $H E X A G O$ is $120^{\circ}$. If $H E=X A=G O=p$ and $E X=A G=O H=q$, compute the area of circle $\omega$.

Solution: The given information implies that triangles $H E X, X A G$, and $G O H$ are congruent, hence triangle $H X G$ is equilateral. If $H X=s$, then the radius of the circle circumscribing $\triangle H X G$ is $s / \sqrt{3}$ so that the circle's area is $\pi s^{2} / 3$. It remains to compute $s$. With $\mathrm{m} \angle H E X=120^{\circ}$, use the Law of Cosines to find

$$
\begin{aligned}
H X^{2} & =H E^{2}+E X^{2}-2 H E \cdot E X \cdot \cos 120^{\circ} \\
& =p^{2}+q^{2}-2 p q(-1 / 2) \\
& =p^{2}+q^{2}+p q .
\end{aligned}
$$

Using the answers 74 and 7 from positions 7 and 9 , respectively, conclude that $S=\{2,7,37\}$ and that $(p, q)=(2,7)$. Hence the foregoing yields $H X^{2}=4+49+14=67$. Thus the area of circle $\omega$ is $\frac{\mathbf{6 7 \pi}}{\mathbf{3}}$.

Final answer: $\frac{67 \pi}{3}$