Determining four rational numbers, given their pairwise sums

Question

The pairwise sums of four weights (rational numbers) are: 6, 8, 10, 12, 15, 16. What are the four weights?

I tried intuitively solving as well as making equations to solve it. If we assume weights in ascending order as: A, B, C, D, then the smallest two weights should sum to 6 (A + B = 6), and the largest two should sum to 16 (C + D = 16).

We can't comment on the order of the rest of the pairs. I tried hit and trial approach with some pairs and working on 4 equations at a time, but I haven't been able to get values which fit all 6 equations.

What would be be a good way to solve this?

Answer 1

Let the weights be $A,B,C,D$ and WLOG assume $A\le B\le C\le D$. You have already concluded that $A+B=6, C+D=16$.

The second smallest sum must be $8=A+C$. Namely, each of the remaining four sums $A+D, B+C, B+D, C+D$ can be easily proven to be $\ge A+C$. Similarly, the second largest sum must be $15=B+D$.

However, this gives us contradiction as $A+B+C+D=(A+B)+(C+D)=6+16=22$ but also $A+B+C+D=(A+C)+(B+D)=8+15=23$.

Therefore, no such weights exist.

Answer 2

More generally, if you have $2n$ weights and the sums of the weights of any $n$ can be written as:

$$w_1\leq w_2\leq \cdots \leq w_{K},$$ where $K=\binom{2n}n,$ then $w_{j}+w_{K+1-j}$ must be constant as $j=1,2,\dots,K.$

In your case, since $w_1+w_6=6+16=22$ and $w_2+w_5=8+15=23,$ there can be no solution.

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Even more generally, if we have $m$ weights $u_1,\dots,u_m$ and a set $J$ of subsets of $\{1,\dots,m\}$ which is closed under complements, and we sort the weights $w_S=\sum_{i\in S} u_i$ for each $S\in J$ as $w_1\leq w_2\leq \cdots \leq w_{|J|}$ we again get that $w_{j}+w_{|J|+1-j}$ must be constant.

Answer 3

$A+B+C+D=6+16=22$ and $$\{A+C,A+D,B+C,B+D\}=\{8,10,12,15\}$$ is impossible, because the four numbers on the RHS cannot be split into two pairs with sum $22.$ The pair containing $15$ will always have an odd sum.

Answer 4

From the conditions, since you assumed that $A_1\leq A_2\leq A_3\leq A_4$, we can conclude that $A_1+A_2=6$, $A_3+A_4=16$, $A_1+A_3=8$ and $A_2+A_4=15$. This gives $A_1+A_2+A_3+A_4=22=23$, which is a contradiction. Hence, it is impossible to satisfy the conditions in the question.

Answer 5

Not a new answer, just an illustration of the arguments concerning even- & oddness of the sum-values in some comments.
Let the unknown base-values, which get pairwise summed denote as $x_1 .. x_4$. Then the six pairwise sums as $s_1...s_6$ . Then let's look at the full multiplication table which occurs when we only look at the residues $(\text{mod } 2)$ (=even- and oddness). We do not find a solution with exactly $1$ odd and $5$ even results for the sums in the complete multiplication table: