Problem 3

There are

4n

pebbles of weights

1, 2, 3, \ldots, 4n

. Each pebble is colored in one of

n

colors and there are four pebbles of each color. Show that we can arrange the pebbles into two piles so that the following two conditions are both satisfied: - The total weights of both piles are the same. - Each pile contains two pebbles of each color.

Step 6 of 6: Verify the two piles balance

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In plain words

Counting blue incidences two different ways shows there are exactly as many blue strings as green strings, and every string weighs the same, so the two piles must weigh the same.

|\text{Pile}_{\text{blue}}|=|\text{Pile}_{\text{green}}|=2n,\qquad \sum \text{Pile}_{\text{blue}}=\sum \text{Pile}_{\text{green}}=n(4n+1).

Detailed analysis

Summing the blue degree over all n vertices counts 2n blue edge-endpoints, and each blue string contributes 2 endpoints, so there are exactly n blue strings; symmetrically there are n green strings, matching the total of 2n strings. Every string weighs 4n+1, so the blue pile weighs n(4n+1) and the green pile weighs n(4n+1) as well, so the two piles have equal weight and, by the previous step, equal colour composition, completing the construction.