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Find the lowest n from N that satisfies:

  1. Oct 20, 2008 #1
    1. The problem statement, all variables and given/known data
    Find the lowest number n from N that satisfies:


    2. Relevant equations

    [​IMG]
    Please see attached
    if k is any number from N


    3. The attempt at a solution
    I have absolutely no idea where to start. :(
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Oct 20, 2008 #2

    tiny-tim

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    Hi Eliva!

    If the attachment is just the equation that you have to prove, then please type it out for us.

    It's not fair on the moderators to make them do the work of checking your attachment when it would be far less work for you to type it out.
     
  4. Oct 20, 2008 #3

    HallsofIvy

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    the attachment says
    a/2= k12
    a/3= k22
    a/5= k32

    Since his statement of the problem say simply "Find the smalllest number from N that satisfies" and ends there, this makes no sense at all.
     
  5. Oct 20, 2008 #4
    Yeah, well, I tried ... the Latex reference was not displaying the equations right, thats why i did it in Word and uploaded an image. Im sorry if that put u into trouble, but the code was not working right, all it gave me in the preview was a few red lines. So again, Im sorry.

    Yes, I might have not explained it quite right. The thing is that u acctually need to find the number a, if all those three statements are correct, which at the same time is suppose to be the lowest (smallest) natural number. As in my knowledge it might have smth to do with finding the GCD of these numbers, or maybe use the Euclidean algorithm, but Im not sure exactly how. k1, k2, and k3 are merely constants, they can be any number, it only means that for instance, in the first statement a/2 exists as a full square from some number K. Please have in mind that all three statements must be correct for the number a.
     
  6. Oct 20, 2008 #5

    HallsofIvy

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    So a= 2k12= 3k22= 5k32 for some numbers k1, k2, k3?

    If the k's and a are required to be integers, there is no such number: the first equation requires that a= sqrt(2) k1 and, since sqrt(2) is not rational, that is impossible.

    I the k's and a are allowed to be general real numbers, there is no smallest such number- just take the k's to be arbitrarily close to 0.
     
    Last edited: Oct 21, 2008
  7. Oct 20, 2008 #6

    tiny-tim

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    Hi HallsofIvy! :smile:

    Well, I can see the .jpg now, but it actually says
    a/2= k12
    a/3= k23
    a/5= k35

    Hi Eliva! :smile:

    So you mean the question is "In the natural numbers, N, find the lowest such that half of it is a perfect square, one-third is a perfect cube, and one fifth is a perfect fifth power"?

    Hint: Start by writing out some equations which combine more than one of the three basic equations (and call them p q and r rather than k1 k2 and k3 … it's easier to type :wink:).

    What do you get? :smile:
     
  8. Oct 20, 2008 #7
    Yessss! thats exactly what Im asking! :) Thank you, my math-english is so bad, Im really having troubles setting the words right.

    not much :) i can see that 2|a, 3|a, 5|a .. but i can't go passed that... i tried doing a more equations with the given statements, but i seem to be going in circle.
    I know I need to use something with divisors or GDC, i just can't see the connection. The GDC(2,3,5)=1, which doesent give me much either.
    I wish there was given at least one more statement or something. Im totaly stuck :(
     
  9. Oct 20, 2008 #8

    tiny-tim

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    Hint: so call it 2a3b5cd :wink:
     
  10. Oct 20, 2008 #9
    ohhh!! i knew it had something to do with those numbers being simple. Ok, it's Euclids algorithm (which i said myself d'uh) .... so the number a can be factorized by simple numbers (2, 3 and 5 in this case) ... but how do I find their powers?
    and just to make sure,
    in here 2a this "a" here is different than the number a we are looking for, right??

    so a= 2x3y5zd
    what is the last digit d there??, would it be 7? is there suppose to be one more multiplier, or i need to only use 2, 3 and 5?? andddd, how do i find the powers (lets call them x,y,z, for no confusion) ??
     
  11. Oct 20, 2008 #10
    Given the way you have written a, what is a/2, a/3, a/5? What must they equal?
     
  12. Oct 21, 2008 #11
    In the natural numbers, N, find the lowest such that half of it is a perfect square, one-third is a perfect cube, and one fifth is a perfect fifth power.

    thats the assignment
     
  13. Oct 21, 2008 #12

    tiny-tim

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    No, Tedjn :smile: meant a= 2x-13y5zd,

    a/3 = … ?

    a/5 = … ? :smile:

    (and d can be any number, not divisible by 2 3 or 5)
     
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