## anonymous one year ago Let n be an odd integer with 11 positive divisors. find the number of positive divisors of 8n^3.

1. ganeshie8

Hey!

2. anonymous

Hello!

3. amistre64

how do we describe 11 positive divisors? say we have a number like 30, what are its positive divisors?

4. amistre64

are they: 2,3,5 ? or, 1,30, 2,15, 3,10, 5,6 ??

5. anonymous

the second set of numbers you gave would be it's positive divisors.

6. anonymous

so number of divisors is tau function right ? are you familiar with it ?

7. anonymous

so when you say n is an odd integer with 11 positive divisors means $$\Large \tau(n)=11$$

8. anonymous

now tau function is multiplicative which means it have this property $$\Large \tau(a\times b)=\tau(a)\times\tau(b)$$

9. anonymous

hello there @contradiction are u with me ?

10. anonymous

so $$\Large \tau(8n^3)=\tau(8)\times(\tau(n))^3$$

11. anonymous

oh!

12. anonymous

sorry for the late response

13. amistre64

dont divisors some in groups of 2?

14. amistre64

hmm, i spose if one set was a perfect square then the grouping a,a would represent a single divisor as opposed to 2 of them

15. amistre64

an algorithm online says that the product of the exponents of the prime factorization ... when you add 1 gives us the number of divisors. so like: n=3^(10) has 11 divisors: (10+1)(0+1)

16. amistre64

n^3 = 3^(30) 2^3 = 8, so (3+1)(30+1) seems to be a specific approach

17. anonymous

ah, right, that makes sense

18. anonymous

okay, i got it! thank you!