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Homework Help: Find smallest value of k in this equation

  1. Jun 12, 2010 #1
    1. The problem statement, all variables and given/known data

    For these equations m,n,p,k are positive whole numbers greater than 1 .

    n^(5/3)=m^(7/2)

    nm=p^k

    What is the smallest value that k can be?

    (A) 6
    (B) 11
    (C) 31
    (D) 41

    2. Relevant equations



    3. The attempt at a solution

    some hints ?
     
  2. jcsd
  3. Jun 12, 2010 #2
    Re: indices

    try to get n in terms of m. Then see what k has to be such that p^k = nm
     
  4. Jun 12, 2010 #3
    Re: indices

    that leaves me with p^k=m^(31/10)

    and i am not told which is greater , p or m
     
  5. Jun 12, 2010 #4
    Re: indices

    you now get m = p^(10k/31) which means p^(10k/31) must be a whole number > 1.
    So what does that mean for (10k/31)?
     
  6. Jun 12, 2010 #5
    Re: indices

    erm 10k/31>0 ?
     
  7. Jun 12, 2010 #6
    Re: indices

    I would suspect that 10k/31 must be an integer so that p^(10k/31) is also an integer. I don't have any proof for this, but someone else might. As such it means that 10k must be a multiple of 31. Does that make sense?

    EDIT: sorry this is incorrect.
     
  8. Jun 12, 2010 #7
    Re: indices

    If we are simply trying to satisfy the equation for the smallest k,
    just solve it like this
    m^(31/10) = p^k => m^(31/10k) = p
    So if k = 6, we just need some m such that its 60th root is a whole number.
    So let m = 2^60 or anything like that.
    we have that n = m^(21/10) so n is also a whole number.

    Does that make sense?
     
  9. Jun 12, 2010 #8
    Re: indices

    well i have another thought on it ,

    m=p^(10k/31)

    For m is a positive whole number, 10k/31 must be an positive integer and p is known to be a positive whole number.

    Therefore ,k must be a multiple of 31 where the smallest is 31 itself. SO the answer is obviously C
     
  10. Jun 12, 2010 #9
    Re: indices

    10k/31 does not need to be a positive integer.
    Consider 9^(1/2)
     
  11. Jun 13, 2010 #10
    Re: indices

    k is known to be a positive whole number from the question and none of the values of k for
    1<k<31 is able to make the indices either a whole number or square roots , cube roots ,...until 31.
     
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