1/p + 1/q + 1/r

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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.
 

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  • #2
Zurtex
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I'm not sure there is a solution.
 
  • #3
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Is there a way of proving this then?
 
  • #4
Hurkyl
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Brute force doesn't work?
 
  • #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.
 
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  • #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.
 
  • #7
Hurkyl
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1/2 +1/4 + 1/5 = 19/20
I can do better!
 
  • #8
Zurtex
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jcsd said:
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.
p=7
q=3
r=2

Is closer
 
  • #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.
 
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  • #10
jcsd
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Hurkyl said:
I can do better!
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.
 
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  • #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.
 
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  • #12
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jcsd said:
2/3 + 2/3 + 1/4 < 1/2 + 1/3 + 1/7 therefore at leats one term muct be a 1/2
What should this equation really read like?
 
  • #13
Zurtex
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recon said:
What should this equation really read like?
I think:

1/3 + 1/3 + 1/4 < 1/2 + 1/3 + 1/7
 

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