Number of Divisor which perfect cubes and multiples of a number n = $2^{14}$$3^9$$5^8$$7^{10}$$11^3$$13^5$$37^{10}$
How many positive divisors that are perfect cubes and multiples of  $2^{10}$$3^9$$5^2$$7^{5}$$11^2$$13^2$$37^{2}$.
I'm able to solve number of perfect square and number of of perfect cubes.
But the extra condition of multiples of  $2^{10}$$3^9$$5^2$$7^{5}$$11^2$$13^2$$37^{2}$ is confusing, anyone can give me a hint?

The numbers you are looking for must be perfect cubes. If you split them into powers of primes, they can have a factor $2^0$, $2^3$, $2^6$, $2^9$ and so on but not $2^1, 2^2, 2^4$ etc. because these are not cubes. The same goes for powers of $3, 5, 7$ and any other primes. 
The numbers must also be multiples of $2^{10}$ so can have factors $2^{12}, 2^{15}, 2^{18}$ etc. because $2^9, 2^6$ and so on are not multiples of $2^{10}$. The numbers must divide $2^{14}$, which leaves only $2^{12}$ because $2^{15}, 2^{18}$ and so on don't divide $2^{14}$. 
You get another factor $3^9, 5^3$ or $5^6, 7^6$ or $7^9, 11^3, 13^3$, and $37^3, 37^6$ or $37^9$. For most powers you have one choice, for $5, 7$ and $11$ you have two choices, for $37$ you have three choices - total is $2 \times 2\times 2\times 3$ numbers. 
