Every positive integer with exactly 91 divisors is a perfect 6-th power???

Discussion in 'General Math' started by don.lotto, Sep 6, 2006.

  1. don.lotto

    don.lotto Guest

    Every positive integer with exactly 91 divisors is a perfect 6-th
    power???

    6th root(7/9 *40!( 6!*34!) ) = ???

    don.mcdonald nz 6/sept 2006.
     
    don.lotto, Sep 6, 2006
    #1
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  2. don.lotto

    don.lotto Guest

    wrote:

    google calc doesnot work any more xx

    please make it work, better.
    factorise etc.

    oucchh!xxx. dear google.

    don.mcdonald.
     
    don.lotto, Sep 6, 2006
    #2
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  3. don.lotto

    Ed Murphy Guest

    Yes, because it must be either (p1^6 * p2^12) or (p1^90).
    2985406 and two-thirds.

    2985406 has four divisors (1, 2, 149203, 2985406).

    2985407 has four divisors (1, 107, 27901, 2985407).

    Why do you expect 7/9 * C(40,6) to be related to a number with 91
    divisors? I don't get it.
     
    Ed Murphy, Sep 6, 2006
    #3
  4. don.lotto

    Ed Murphy Guest

    Google Calculator handles factorials just fine:

    (7 / 9) * ((40 !) / ((6 !) * (34 !))) = 2 985 406.67

    ((7 / 9) * ((40 !) / ((6 !) * (34 !))))^(1 / 6) = 11.9996133

    Perhaps you botched the parentheses? I did, the first time.

    This is indeed close to 2^12 * 3^6 = 2985984 (a difference of 577 1/3).
     
    Ed Murphy, Sep 6, 2006
    #4
  5. Yes. 91 = 7*13, so any number with 91 divisors is of the
    form p^6 q^12 or r^90.
     
    Pubkeybreaker, Sep 6, 2006
    #5
  6. don.lotto

    don.lotto Guest

    well, 6th root didnot kick in the calc.

    sometimes 3*5 does not kick in the calc.
    something like that.

    ouch 40! / (...) xxxxxxxxx division may be missing.

    ((7 / 9) * (40 choose 6))^(1 / 6) = 11.9996133
    More about calculator.


    thanks Ed. and pub key.

    It seemed to be much freer with respect to parentheses in the past.

    i will keep trying. but it haS BEEN harder and less useable lately.
    don.mcdonald.
     
    don.lotto, Sep 7, 2006
    #6
  7. don.lotto

    don.lotto Guest

    ....
    Re: Every positive integer with exactly 91 divisors is a perfect 6-th
    power???

    ((7 / 9) + (1 / 6 648)) * (40 choose 6) = 2 985 984.04

    a 7-digit phone number. (04) = wellington and kapiti new zealand.

    = 12^ 6th.
    the smallest positive integer with 91 divisors.

    I am struggling with calc google.


    greater than.
    An estimate of its magnitude. cheers,

    don.mcdonald.
    author's solution soon.
     
    don.lotto, Sep 7, 2006
    #7
  8. don.lotto

    bill.daly Guest

    Exactly, but it is possible to improve on this. If d is a product of
    primes, each of which is of the form 6n+1, then any number with exactly
    d divisors is a 6th power. For that matter, if m is any number, and d
    is a product of primes, each of which leaves a residue of 1 mod m, then
    any number with exactly d divisors is an mth power.
     
    bill.daly, Sep 13, 2006
    #8
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