Positive divisors of n = $2^{14} \cdot 3^9 \cdot 5^8 \cdot 7^{10} \cdot 11^3 \cdot 13^5 \cdot 37^{10}$ How do I find positive divisors of n that are perfect cubes that are multiples of 2^10 * 3^9 * 5^2 * 7^5 * 11^2 * 13^2 * 37^2
The answer is (1)(1)(2)(2)(1)(1)(3) = 12
I don't understand though because I would have done something like:
2: [(14-10)/3]+1 = 2 (taking the floor)
3: [(9-9)/3]+1 = 1
5: [(8-2)/3]+1 = 3
7: [(10-5)/3]+1 = 2
11: [(3-2)/3]+1 = 1
13: [(5-2)/3]+1 = 2 
37: [(10-2)/3]+1 =3
2*1*3*2*1*2*3  

For $d$ to be a divisor of $n$, $d$ must be of the form $2^a \cdot 3^b \cdot 5^c \cdot 7^d \cdot 11^e \cdot 13^f \cdot 37^g$, where $0 \leq a \leq 14, 0 \leq b \leq 9, 0 \leq c \leq 8, 0 \leq d \leq 10, 0 \leq e\leq 3, 0\leq f \leq 5, 0 \leq g \leq 10$.
Now we want $d$ ot be a multiple of the number given, that means $d$ must be of the form $2^a \cdot 3^\color{red}{9} \cdot 5^c \cdot 7^d \cdot 11^e \cdot 13^f \cdot 37^g$, where $\color{red}{10} \leq a \leq 14, \color{red}{2} \leq c \leq 8, \color{red}{5} \leq d \leq 10, \color{red}{2} \leq e\leq 3, \color{red}{2}\leq f \leq 5, \color{red}{2} \leq g \leq 10$.
Now we want $d$ to be a cube as well. This means all powers appearing must by divisible by $3$.  Thus
$$a=12, b=9, c \in \{3,6\}, d \in \{6,9\}, e=3, f=3, g \in \{3,6,9\}.$$
Thus the total number of choices we have  are
$$2 \cdot 2 \cdot 3=12.$$
