Sum of Weights From Pairwise Sums
You have five objects with distinct weights. When weighed in all possible pairs of two, the pairwise sums are: $$31, 32, 33, 34, 35, 36, 37, 38, 39, 40 \text{ kg}$$
What is the total weight of all five objects?
Hints
- Don't try to solve for individual weights -- there is a shortcut that gives you the total directly.
- How many of the 10 pairwise sums does each individual weight appear in? Use this multiplicity to relate the sum of all pairwise sums to the total weight.
- Each of the 5 weights appears in exactly 4 pairwise sums, so $\sum \text{(pairwise sums)} = 4(a+b+c+d+e)$. Compute $31 + 32 + \cdots + 40 = 355$ and divide by 4.
Worked Solution
How to Think About It: You have 5 unknowns and are given 10 pairwise sums. Solving a system of 10 equations in 5 unknowns is overkill if all you need is the total. The key observation is a symmetry argument: each weight appears in the same number of pairwise sums. If you add up all the pairwise sums, each individual weight gets counted the same number of times, so you can recover the total directly.
Quick Estimate: The 10 pairwise sums range from 110 to 121, with mean $115.6$. Each pairwise sum averages two weights, so the average individual weight is about $115.6 / 2 \approx 57.8$. The total of five weights is roughly $5 \times 57.8 = 289$. Let's see if the exact answer is close.
Approach: Count how many times each weight appears across all pairwise sums, then relate the sum of pairwise sums to the total weight.
Formal Solution:
Label the five weights $a, b, c, d, e$. The $\binom{5}{2} = 10$ pairwise sums are: $$a+b,\; a+c,\; a+d,\; a+e,\; b+c,\; b+d,\; b+e,\; c+d,\; c+e,\; d+e$$
Each weight is paired with the other 4 weights, so each weight appears in exactly 4 of the 10 sums. Adding all pairwise sums: $$\sum_{\text{all pairs}} = 4(a + b + c + d + e) = 4S$$
where $S = a + b + c + d + e$ is the total weight.
The sum of the given pairwise values is: $$110 + 112 + 113 + 114 + 115 + 116 + 117 + 118 + 120 + 121 = 1156$$
So: $$4S = 1156 \implies S = \frac{1156}{4} = 289$$
Answer: The total weight of all five objects is $\dfrac{1156}{4} = 289$ kg.
Intuition
This is a beautiful example of how symmetry can collapse a seemingly complex system into a single equation. Instead of solving for 5 individual unknowns, you notice that every weight plays the same structural role -- each one appears in exactly 4 of the 10 pairwise sums. So the sum of all pairwise sums is exactly 4 times the total weight. This double-counting argument is one of the most powerful tricks in combinatorics.
The same principle applies broadly: whenever you have symmetric aggregates of unknowns, adding them all up often produces a clean multiple of the quantity you want. In finance, this shows up when you have pairwise correlations or covariances and want to recover portfolio-level quantities. The lesson is always to ask "what happens when I add everything up?" before reaching for a system of equations.