[Number Theory]
Problem: Let $k$ be the least common multiple of the numbers in the set $\mathcal{S}=\{1,2, \ldots, 30\}$. Determine the number of positive integer divisors of $k$ that are divisible by exactly 28 of the numbers in the set $\mathcal{S}$.

Solution: We know that $k=2^{4} \cdot 3^{3} \cdot 5^{2} \cdot 7 \cdot 11 \cdot 13 \cdot 17 \cdot 19 \cdot 23 \cdot 29$. It is not difficult to see that the set $\mathcal{T}_{1}=\left\{\frac{k}{2}, \frac{k}{3}, \frac{k}{5}, \frac{k}{17}, \frac{k}{19}, \frac{k}{23}, \frac{k}{29}\right\}$ comprises all divisors of $k$ that are divisible by exactly 29 of the numbers in the set $\mathcal{S}$. Let $\mathcal{P}=\{2,3,5,17,19,23,29\}$. Then

$$
\mathcal{T}_{2}=\left\{\frac{k}{p_{1} p_{2}}, \text { where } p_{1} \text { and } p_{2} \text { are distinct elements of } \mathcal{P}\right\}
$$

consists of divisors of $k$ that are divisible by exactly 28 of the numbers in the set $\mathcal{S}$. There are $\left(\begin{array}{l}7 \\ 2\end{array}\right)=21$ elements in $\mathcal{T}_{2}$.

Furthermore, note that $\frac{k}{7}$ is only divisible by 26 of the numbers in $\mathcal{S}$ (since it is not divisible by $7,14,21$, or 28 ) while $\frac{k}{11}$ and $\frac{k}{13}$ are each divisible by 28 of the numbers in $\mathcal{S}$. We can also rule out $\frac{k}{4}$ (27 divisors: all but 8,16 , and 24 ), $\frac{k}{9}$ (27 divisors), $\frac{k}{25}$ (24 divisors), and all other numbers, thus the answer is $21+2=\mathbf{2 3}$.

Final answer: 23