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Math Help - positive divisor

  1. #1
    rcs
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    positive divisor

    Let n = 2^31 3^19. How many positive divisors of n^2 are less than n but
    do not divide n?
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  2. #2
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    Quote Originally Posted by rcs View Post
    Let n = 2^31 3^19. How many positive divisors of n^2 are less than n but
    do not divide n?

    Lemma: if n=p_1^{a_1}\cdot\ldots\cdot p_k^{a_k} is the prime decomposition of a natural number n, then the number of positive divisors of n is \prod\limits_{i=1}^k(a_i+1)

    Well, now apply the above both to n\,\,and\,\,n^2 and do some maths...


    Tonio
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    Quote Originally Posted by rcs View Post
    Let n = 2^{31}\cdot 3^{19}. How many positive divisors of n^2 are less than  n but
    do not divide  n?
    A divisor of n^2 must be of the form 2^p\cdot3^q, with 0\leqslant p\leqslant 62 and 0\leqslant q\leqslant 38. If it does not divide  n then either p>31 or q>19.

    Now comes the tricky part. The remaining condition is that 2^p\cdot3^q < n. I suggest that you investigate this by choosing a random value of p between 31 and 62, say p=45, and looking at how many values of q satisfy the condition 2^{45}\cdot 3^q <2^{31}\cdot 3^{19}. [Hint: take logs.] That may possibly give you some insight into how to tackle the problem in general.
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  4. #4
    rcs
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    it is said that the answer is 589 but i dont know how it is reached to this.
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    This question has already been discussed, but I have not tried to find a more elegant solution.
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