Aerith picks five numbers and for every three of them, takes their product, producing ten products. She tells Bob that the nine smallest positive divisors of sixty are among her products. Can Bob figure out the last product?
Yes Bob can. Let the numbers be $v, w, x, y, z$ and let the missing product be $x y z$. The nine given products are $1,2,3,4,5,6,10,12,15$.

Partition these products into the three sequences

$$
s_{1}=(v y z, v z x, v x y), s_{2}=(w y z, w z x, w x y), s_{3}=(v w x, v w y, v w z)
$$

and call these sequences special. Note that $s^{\prime}=(y z, z x, x y)$ is directly proportional to the first two and inversely proportional to the last one. The multiples of 5 thus have to be in the same special sequence, WLOG $(5,10,15)$. This sequence is inversely proportional to at least one of other two special sequences, which would thus have to be a multiple of $(6,3,2)$. However, it cannot be $(6,3,2)$ itself as that would leave 1,4 , and 12 , which cannot form a special sequence. Thus, the multiple of $(6,3,2)$ has to be $(12,6,4)$, leaving $(1,2,3)$ as the final sequence. WLOG, we thus can get the equations

$((v y z, v z x, v x y),(w y z, w z x, w x y),(v w x, v w y, v w z))=((5,10,15),(1,2,3),(12,6,4))$.

We can then deduce $x y z=v y z \cdot w x z / v w z=5 \cdot 2 / 4=5 / 2$.

Indeed, $5 / 2$ is achieved by

$$
(v, w, x, y, z)=\left(2 \sqrt[3]{15}, \frac{2}{5} \sqrt[3]{15}, \sqrt[3]{15}, \frac{1}{2} \sqrt[3]{15}, \frac{1}{3} \sqrt[3]{15}\right)
$$