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1/p + 1/q + 1/r

  1. Jan 10, 2005 #1
    What is the maximum sum less than 1 but more than 0 that can be formed from [tex]\frac{1}{p} + \frac{1}{q} + \frac{1}{r}[/tex], where p, q and r are positive integers?

    I posted this problem in the homework forum even though it isn't homework. Since I have not yet received any help, I have posted it here.
     
  2. jcsd
  3. Jan 10, 2005 #2

    Zurtex

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    I'm not sure there is a solution.
     
  4. Jan 10, 2005 #3
    Is there a way of proving this then?
     
  5. Jan 10, 2005 #4

    Hurkyl

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    Brute force doesn't work?
     
  6. Jan 10, 2005 #5

    jcsd

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    1/2 +1/4 + 1/5 = 19/20

    edited to add: yes brute force is the way, but you onluy need actually consider about 5 or so different sums once you've excluded the ones thta cannot possibly be the correct answer. For example the fact that 1/2 + 1/4 + 1/5 > 1/4 + 1/4 + 1/4 implies that at least one term must be greater than 1/4.
     
    Last edited: Jan 10, 2005
  7. Jan 10, 2005 #6

    Zurtex

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    If you rearrange this you can show you want pq + qr + rp to be as close as possible to pqr but such that pq + qr + rp < pqr. Can't say my maths is good enough to think about 3 independent variables like this.

    But it would seem to me if you could up with some functions p(t), q(t) and r(t) such that p(t)q(t) + q(t)r(t) + r(t)p(t) monotonically increases and so does p(t)q(t)r(t) and that p(t)q(t) + q(t)r(t) + r(t)p(t) < p(t)q(t)r(t) holds true for all [itex] p(t), q(t), r(t) \in \mathBB{N}[/itex] then that is proof there is no solution.

    Not very useful I know I'm sorry.
     
  8. Jan 10, 2005 #7

    Hurkyl

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    I can do better!
     
  9. Jan 10, 2005 #8

    Zurtex

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    p=7
    q=3
    r=2

    Is closer
     
  10. Jan 10, 2005 #9

    Gokul43201

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    How about [tex] \frac{1}{2} + \frac{1}{3} + \frac{1}{7}~~? [/tex]

    Edit : didn't see zurtex's post.
     
    Last edited: Jan 10, 2005
  11. Jan 10, 2005 #10

    jcsd

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    hmm lets try again:
    oops yes, I actually did consider: 1/2 + 1/3 + 1/7 = 41/42, but I added it up wrong:blushing:

    edited to add:

    Here's the proof then tthat this is indeedf the largets possible answer.

    As I said earlier we already know that at least one term must be greater than 1/4:

    2/3 + 2/3 + 1/4 < 1/2 + 1/3 + 1/7 therefore at leats one term muct be a 1/2

    As 1/2 + 1/3 + 1/6 > 1 and 1/2 + 1/3 + 1/7 < 1, the rest of the terms must be less than or equal to 1/3 and gretare than or equal to 1/7

    which means only the sums:

    1/2 + 1/3 + 1/7

    1/2 + 1/4 + 1/5

    1/2 + 1/6 + 1/6

    need be considered of them the greaest is 1/2 + 1/3 + 1/7.

    edited again to add: looks like Gokul beat me to it.
     
    Last edited: Jan 10, 2005
  12. Jan 10, 2005 #11

    Gokul43201

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    Re-write the problem as follows :
    For a given value of r, find p, q that maximizes the sum of reciprocals subject to it being less than 1.

    If r> 7, need maximal 1/p + 1/q < (r-1)/r.

    This is achieved with p=2, q=3 for all such r, since :
    1/2 + 1/3 < 7/8 => 1/2 + 1/3 < (r-1)/r , for all r>7
    AND
    the next larger value of 1/p + 1/q = 1/2 + 1/2, which is clearly not allowed, as the sum will exceed 1.

    Thus the maximal solution for any r>7 is 1/2, 1/3, 1/r

    Clearly, any of these solutions is inferior to 1/2, 1/3, 1/7.

    This leaves you with only having to check r=5,6.
     
    Last edited: Jan 10, 2005
  13. Jan 10, 2005 #12
    What should this equation really read like?
     
  14. Jan 11, 2005 #13

    Zurtex

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    I think:

    1/3 + 1/3 + 1/4 < 1/2 + 1/3 + 1/7
     
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