anonymous
  • anonymous
Let n be an odd integer with 11 positive divisors. find the number of positive divisors of 8n^3.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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ganeshie8
  • ganeshie8
Hey!
anonymous
  • anonymous
Hello!
amistre64
  • amistre64
how do we describe 11 positive divisors? say we have a number like 30, what are its positive divisors?

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amistre64
  • amistre64
are they: 2,3,5 ? or, 1,30, 2,15, 3,10, 5,6 ??
anonymous
  • anonymous
the second set of numbers you gave would be it's positive divisors.
anonymous
  • anonymous
so number of divisors is tau function right ? are you familiar with it ?
anonymous
  • anonymous
so when you say n is an odd integer with 11 positive divisors means \(\Large \tau(n)=11\)
anonymous
  • anonymous
now tau function is multiplicative which means it have this property \(\Large \tau(a\times b)=\tau(a)\times\tau(b)\)
anonymous
  • anonymous
hello there @contradiction are u with me ?
anonymous
  • anonymous
so \(\Large \tau(8n^3)=\tau(8)\times(\tau(n))^3\)
anonymous
  • anonymous
oh!
anonymous
  • anonymous
sorry for the late response
amistre64
  • amistre64
dont divisors some in groups of 2?
amistre64
  • 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
amistre64
  • 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)
amistre64
  • amistre64
n^3 = 3^(30) 2^3 = 8, so (3+1)(30+1) seems to be a specific approach
anonymous
  • anonymous
ah, right, that makes sense
anonymous
  • anonymous
okay, i got it! thank you!

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