[Combinatorics]
Problem: Let $T=5$. Compute the number of positive divisors of the number $20^{4} \cdot 11^{T}$ that are perfect cubes.

Solution: Let $N=20^{4} \cdot 11^{T}=2^{8} \cdot 5^{4} \cdot 11^{T}$. If $m \mid N$, then $m=2^{a} \cdot 5^{b} \cdot 11^{c}$ where $a, b$, and $c$ are nonnegative integers such that $a \leq 8, b \leq 4$, and $c \leq T$. If $m$ is a perfect cube, then $a, b$, and $c$ must be divisible by 3 . So $a=0,3$, or $6 ; b=0$ or 3 , and $c \in\{0,3, \ldots, 3 \cdot\lfloor T / 3\rfloor\}$. There are a total of $3 \cdot 2 \cdot(\lfloor T / 3\rfloor+1)$ possible values of $m$. For $T=5,\lfloor T / 3\rfloor+1=2$, so the number of possible values of $m$ is $\mathbf{1 2}$.

Final answer: 12