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Find five numbers

  1. Oct 13, 2008 #1

    thrill3rnit3

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    1. The problem statement, all variables and given/known data

    Find five numbers for which the pair sums are equal to 0,2,4,5,7,9,10,12,14,17

    2. Relevant equations

    nothing.

    3. The attempt at a solution

    First, pair sums means sum of 2 numbers, right?

    I just need a heads start on this one. I don't know how to get started. Just give hints please. No full solutions.

    Thanks.
     
  2. jcsd
  3. Oct 13, 2008 #2

    Borek

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    x1+x2=0
    x1+x3=2
    ...

    Looks like a system of linear equations to me. xi are your unknowns.
     
    Last edited: Oct 13, 2008
  4. Oct 13, 2008 #3

    thrill3rnit3

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    I don't think so, if it were

    [tex]x_{1}+x_{n}=y[/tex]

    then I could just substitute any number for the x1, and it would produce 10 numbers, plus 1 for the for the x1, which makes it 11. The problem asks for only 5 numbers.
     
  5. Oct 13, 2008 #4

    Borek

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    Write all combinations of two unknowns. These are not only x1+something.
     
  6. Oct 13, 2008 #5

    thrill3rnit3

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    Yeah, I've written them all down. I don't see the point to it though. Please elaborate a little bit more.
     
  7. Oct 14, 2008 #6

    Borek

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    How many equations in five unknowns did you get?
     
  8. Oct 14, 2008 #7

    thrill3rnit3

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    I got ten equations (thus the 10 sums), but I'm not really sure how to go from here since I can't just plug in the sums because I'm not sure to which equation each one would go into.
     
  9. Oct 14, 2008 #8

    thrill3rnit3

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    i got the following equations out of it:

    a+b+c+d+e=20

    a+b+2c+2d+2e=23

    3a+3b+4c+4d+4e=63

    3a+3b+5c+5d+5e=66

    which led to

    a+b=17

    c+d+e=3

    I'm not really sure how to do systems in 4 variables. I'll probably work on it right now.
     
    Last edited: Oct 14, 2008
  10. Oct 14, 2008 #9

    Borek

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    My idea of simple systematic approach was wrong, sorry about that.
     
  11. Oct 14, 2008 #10

    thrill3rnit3

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    Yeah that's fine.

    I'm not really fluent with Linear Algebra and matrices and stuff like that. Is there any way to solve the system without using matrices?
     
  12. Oct 14, 2008 #11

    thrill3rnit3

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    So using systems of equations is not applicable on this one??

    By the way, I'm not sure about the 2nd equation I got. I subtracted all the expressions and pair sums to arrive to the 2nd equation, but I forgot that subtraction is NOT commutative.

    I subtracted 17-14-12-10... and got -46

    However, when I tried 0-2-4-5... I ended up with -80

    I'm not really sure how to go about with this.
     
  13. Oct 14, 2008 #12
    Are we assuming that the solution are integers?
     
  14. Oct 14, 2008 #13

    thrill3rnit3

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    I suppose so. I copied down the original problem word for word (see first post), so I would guess the answers would be positive/negative integers
     
  15. Oct 14, 2008 #14

    Dick

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    Assume the numbers are in order a<=b<=c<=d<=e. You did a great trick in getting a+b+c+d+e=20. You counted combinations, right? Brilliant. There are four other equations that are pretty clear. a+b=0, a+c=2, d+e=17 and c+e=14. Do you see why? That's five equations in five unknowns. You can solve them. Borek (and I) underestimated this problem when we first saw it. There IS a systematic approach. More complicated than I thought.
     
    Last edited: Oct 15, 2008
  16. Oct 15, 2008 #15
    I have one trick that can narrow down the solution. There are five numbers right. The possible combinations of the parity of these numbers are

    1. all odd
    2. 1 even, 4 odd

    and so on

    However, we see that in the set of pairwise sums that are given, there are only 4 odd numbers. This shows that the solution can be of only the following two types (can be easily seen)

    1. 1 even, 4 odd
    2. 1 odd, 4 even

    other combinations will give a different set of pairwise sums.

    However, since we know a+b+c+d+e = 20. the solution must be of the form 1 even, 4 odd.

    Say, the even number is a

    then the pairwise sums of a, must be

    a + b = 5
    a + c = 7
    a + d = 9
    a + e = 17

    thats all i got till now.


    I don't think there is a systematic way as you have given. Since we do not know which set of numbers give which sums.

    as u have said, if we assume a+b = 0, how can you be sure that it is the same a involved in a + c = 2??? you could have two other numbers that give sum 2.

    remember that the solution can be negative... they are not all positive. Assuming a<b<c<d<e, does not allow us two create algebraic equations as you have done
     
    Last edited: Oct 15, 2008
  17. Oct 15, 2008 #16
    ok i got the solution......here it is



    From my earlier post we know

    a+b=5
    a+c=7
    a+d=9
    a+e=17

    Also a+b+c+d+e=20

    Subtract the last one from the others. This gives

    c+d+e=15
    b+d+e=13
    b+c+e=11
    b+c+d=3

    Add all of these

    b+c+d+e = 14

    Thus a = 6 (a+b+c+d+e = 20, remember??)

    Put the same a in the first set of equations. This gives

    a=6, b=-1, c=1, d=3, e=11

     
  18. Oct 15, 2008 #17

    Borek

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    Let's assume b+c is the second lowest sum - that means a+c>b+c. But we know that a<b, so it can't be true. Similar happens for every other pair (a+c>b+d, but b<a && d<c, and so on). Thus a+c must be the second lowest sum of all.
     
  19. Oct 15, 2008 #18
    i see your point....

    then we know that a+b = 0 and a+c = 2. However, we can't make anymore such assumptions. can we???

    I mean, the next smallest sum could be a+d or b+c. How do we make sure which it is?
     
  20. Oct 15, 2008 #19

    Borek

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    We don't. But we have two other equations on the other end, that (combined with total sum being 20, which I must admit I have not yet understood :grumpy:) gives five equations in five unknowns.
     
  21. Oct 15, 2008 #20

    thrill3rnit3

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    So do I just subtract the expression from a+b+c+d+e=20??

    e.g.

    (a+b+c+d+e=20)-(a+b=5)

    etc. and so on and so forth? I think that makes sense...
     
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