南瓜配对重量

设 5 个南瓜重量从轻到重为 $a,b,c,d,e$。

若把所有两两配对的重量求和，则每个南瓜会在 4 个配对里出现一次，因此

$$4(a+b+c+d+e)=21+22+\cdots+30=255$$

从而

$$a+b+c+d+e=\frac{255}{4}.$$

补充：原题给出的配对集合在严格意义下实际上无法由 5 个实数重量实现（不存在一组 5 个数使其 10 个两两和恰好为 21..30）。上述结果是按“配对和必须等于 21..30”这一条件直接推出的总和。

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Original Explanation

Let $a, b, c, d,$ and $e$ be the weight of each pumpkin from lightest to heaviest. If we sum all the pairwise weights, we know we will have 4 sets of $a, b, c, d,$ and $e$. Thus

$$4 \cdot (a+b+c+d+e) = 21 + \dots + 30 = 255$$

or

$$a + b + c + d + e = \frac{255}{4}$$

Note: You can't create equations for all the pairs to find the weights of each pumpkin, because it's impossible! There is no set of $5$ numbers for which their pairwise sums complete the set $[21, \dots, 30]$. Big thanks to @.cutnpaste for pointing this out!