Borek said:My idea of simple systematic approach was wrong, sorry about that.
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 don't think there is a systematic way as you have given. Since we do not know which set of numbers give which sums.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.
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
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.
praharmitra said:I mean, the next smallest sum could be a+d or b+c. How do we make sure which it is?
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
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
Dick said:You CAN TOO create the equations systematically.