Let n be an odd integer with 11 positive divisors. find the number of positive divisors of 8n^3.

Question

ganeshie8 · Answer

Hey!

anonymous · Answer

Hello!

amistre64 · Answer

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

amistre64 · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

hello there @contradiction  are u with me ?

anonymous · Answer

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

anonymous · Answer

oh!

anonymous · Answer

sorry for the late response

amistre64 · Answer

dont divisors some in groups of 2?

amistre64 · Answer

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

amistre64 · Answer

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)

amistre64 · Answer

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

anonymous · Answer

ah, right, that makes sense

anonymous · Answer

okay, i got it! thank you!