Solving 5-Pair Sums Equal to 0,2,4,5,7,9,10,12,14,17

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Homework Help Overview

The problem involves finding five numbers such that their pairwise sums equal a specified set of values: 0, 2, 4, 5, 7, 9, 10, 12, 14, and 17. The context is rooted in algebra and systems of equations.

Discussion Character

  • Exploratory, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss the interpretation of pair sums and the implications of forming equations from them. Some suggest writing out combinations of sums, while others express uncertainty about how to assign sums to specific equations. There are attempts to derive relationships between the unknowns and the sums.

Discussion Status

The discussion is ongoing, with various approaches being explored. Some participants have proposed equations based on their interpretations of the sums, while others question the validity of these assumptions. There is a recognition of the complexity involved in solving the system, and some participants have offered insights into the nature of the numbers, such as parity considerations.

Contextual Notes

There is a discussion about whether the solutions are restricted to integers and the implications of this assumption. Participants also consider the ordering of the numbers and how it affects the equations formed from the pairwise sums.

thrill3rnit3
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Homework Statement



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

Homework Equations



nothing.

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.
 
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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:
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.
 
Write all combinations of two unknowns. These are not only x1+something.
 
Yeah, I've written them all down. I don't see the point to it though. Please elaborate a little bit more.
 
How many equations in five unknowns did you get?
 
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.
 
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:
My idea of simple systematic approach was wrong, sorry about that.
 
  • #10
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?
 
  • #11
Borek said:
My idea of simple systematic approach was wrong, sorry about that.

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.
 
  • #12
Are we assuming that the solution are integers?
 
  • #13
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
 
  • #14
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.
 
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  • #15
praharmitra said:
Are we assuming that the solution are integers?

thrill3rnit3 said:
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

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.

Dick said:
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.
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:
  • #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

 
  • #17
praharmitra said:
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

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.
 
  • #18
Borek said:
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.

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?
 
  • #19
praharmitra said:
I mean, the next smallest sum could be a+d or b+c. How do we make sure which it is?

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 ) gives five equations in five unknowns.
 
  • #20
praharmitra said:
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

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...
 
  • #21
praharmitra said:
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

You CAN TOO create the equations systematically. The lowest sum is 0=a+b, the next lowest is 2=a+c, the highest must be d+e=17, the second highest must be c+e=14. Combining this with the equation derived from the sum of all pairs a+b+c+d+e=20 gives the same solution you've given, e.g. subtract a+b=0 and d+e=17 from the pairs equation gives c=3. Since c+e=14, that gives e=11. Etc. The parity trick is clever but unnecessary.
 
  • #22
Dick said:
You CAN TOO create the equations systematically.

Exactly!
 

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