Track / Stage 6 / 55 of 400 #1055 of 1964

Problem 1055

National olympiad, first round

Combinatorics Difficulty 6.1 Prove it

## [Examples and Counterexamples. Constructions]

a) Is it possible to number the edges of a cube with natural numbers from 1 to 12 so that for each vertex of the cube, the sum of the numbers of the edges that meet at it is the same?

b) The same question if the edges of the cube are to be labeled with the numbers $-6,-5,-4,-3,-2,-1,1,2,3,4,5,6$ .

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

a) Suppose this is possible, and the sum of the numbers of the edges converging at each of the eight vertices is $x$ . Adding these sums for all vertices, we get $8x$ . On the other hand, this sum is equal to twice the sum of the numbers of all edges, since the number of each edge appears in it twice. Let's calculate this sum: $2 \cdot (1 + 2 + \ldots + 12) = 156$ . From this, it follows that $8x = 156$ , which means $x$ is not an integer. Contradiction.

b) See the figure:

## Answer

a) It is not possible; b) it is possible.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.