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Math Help - difficult problem

  1. #1
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    difficult problem

    Let "n= 2^{31}3^{19}. How many positive divisors of n^2 are less than n but
    do not divide n?

    the answer is 589

    please give me the complete solution....
    thanks in advance
    Last edited by jpmath2010; November 30th 2010 at 03:51 AM.
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  2. #2
    MHF Contributor

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    "Raise" is an odd notation! Powers are commonly written as "^". I think you mean "n= 2^{31}3^{19}" or, better, using LaTex, " n= 2^{31}3^{19}". n^2 then is n^2= 2^{62}3^{38}. To be a divisor of that but not of n itself, a number must be divisible by either 2^{32} or 3^{20}.

    Now, what additional factors can you add and still keep the number less than n?
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