Problem (Number Theory):
Let $T=T N Y W R$. Compute the number of positive perfect cubes that are divisors of $(T+10) !$.

Solution:
Let $N=T+10$. In order for $k^{3}(k \in \mathbb{N})$ to be a divisor of $N$ !, the largest odd prime factor of $k$ (call it $p$ ) must be less than or equal to $N / 3$ so that there are at least three multiples of $p$ among the product of the first $N$ positive integers. If $p=3$, then the smallest possible value of $N$ is 9 , and the largest perfect cube factor of 9 ! is $2^{6} \cdot 3^{3}$. Similarly, if $p=5$, then the smallest possible value of $N$ is 15 , and the largest perfect cube factor of 15 ! is $2^{9} \cdot 3^{6} \cdot 5^{3}$. With $T=7, N=17$, and the largest perfect cube factor of 17 ! is $2^{15} \cdot 3^{6} \cdot 5^{3}$. Thus $k^{3} \mid 17$ ! if and only if $k \mid 2^{5} \cdot 3^{2} \cdot 5^{1}$. Therefore $k=2^{x} 3^{y} 5^{z}$, where $x, y, z$ are nonnegative integers with $x \leq 5, y \leq 2, z \leq 1$, yielding $6 \cdot 3 \cdot 2=\mathbf{3 6}$ possible values of $k$.

Final answer: 36