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anonymous

  • one year ago

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

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  1. ganeshie8
    • one year ago
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    Hey!

  2. anonymous
    • one year ago
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    Hello!

  3. amistre64
    • one year ago
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    how do we describe 11 positive divisors? say we have a number like 30, what are its positive divisors?

  4. amistre64
    • one year ago
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    are they: 2,3,5 ? or, 1,30, 2,15, 3,10, 5,6 ??

  5. anonymous
    • one year ago
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    the second set of numbers you gave would be it's positive divisors.

  6. anonymous
    • one year ago
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    so number of divisors is tau function right ? are you familiar with it ?

  7. anonymous
    • one year ago
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    so when you say n is an odd integer with 11 positive divisors means \(\Large \tau(n)=11\)

  8. anonymous
    • one year ago
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    now tau function is multiplicative which means it have this property \(\Large \tau(a\times b)=\tau(a)\times\tau(b)\)

  9. anonymous
    • one year ago
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    hello there @contradiction are u with me ?

  10. anonymous
    • one year ago
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    so \(\Large \tau(8n^3)=\tau(8)\times(\tau(n))^3\)

  11. anonymous
    • one year ago
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    oh!

  12. anonymous
    • one year ago
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    sorry for the late response

  13. amistre64
    • one year ago
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    dont divisors some in groups of 2?

  14. amistre64
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    n^3 = 3^(30) 2^3 = 8, so (3+1)(30+1) seems to be a specific approach

  17. anonymous
    • one year ago
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    ah, right, that makes sense

  18. anonymous
    • one year ago
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    okay, i got it! thank you!

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