# 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?

-
Are you asking how many perfect cube divisors n has? – Ethan Dec 16 '12 at 4:10
@Ethan Yes, im need to find out how many perfect cube divisors are multiples of (2^10)(3^9)(5^2)(7^5)(11^2)(13^2)(37^2). – Kai Dec 16 '12 at 4:15
I don't understand, if its a divisor of that number, it cant also be a multiple of it – Ethan Dec 16 '12 at 4:16
@Ethan im able to find out the perfect cube divisor of the n, now, i want check how many of the perfect cube divisors are multiples of (2^10)(3^9)(5^2)(7^5)(11^2)(13^2)(37^2), it is different with the n . – Kai Dec 16 '12 at 4:22
Ok I understand hold on – Ethan Dec 16 '12 at 4:25

Let your first number be n, your second one be j, $5*7*11*13*37*2^2*j$ gives you your first cube that divides n and is a multiple of j, if you multiply this cube by any prime combination 2,3,5,7,11,13,37 cubed, you will get another cube that will also divide n, as long as its exponents don't exceed that of n's, so subtract the corresponding exponents of n and j, divide each number you get by 3, now take the closest integer greater then that number, and keep that, now with each number youve gotten multiply them all together, this should be your anwser, as it should generate all the different combination of possible prime cubes you could multiply to j to get a cube divisor. Your answer is $2^3*3^2$, I think. (Might have made a mistake, but the reasoning to the solution should be very similar)

-
Thank you, sometimes simple things is quite confusing at the beginning. – Kai Dec 16 '12 at 5:00
Im not sure if my answer is correct, im pretty sure though, I don't have any background on combinatorics, so its mostly intuition. – Ethan Dec 16 '12 at 5:02
ohh, actually i have the final answer already, just cant reach there. The answer is (1)(1)(2)(2)(1)(1)(3)=12. Obtained by find the different of numbers of perfect cube divisors between n and the multiply number. (figured out from your explanation but still not fully understand, however obtain answer) – Kai Dec 16 '12 at 5:24

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.

-
If you put dollars signs around your numbers, they format more nicely. Also it will render nicer +-= signs ($+-=$), variables will be in italics, and ^ will work correctly. This is the basic workings of LaTeX, two others which are useful is \frac{ top }{ bottom } to get fractions (e.g. $\frac{top}{bottom}$) and \times to get a nice multiplcation sign (e.g. $2\times 2$ instead of $2x2$) – Eric Stucky Mar 22 '14 at 9:09