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It is not possible. In proving this, we call the sum of the two
numbers at the ends of an edge the 'weight' of that edge. Note that
three of the nodes are connected to exactly two nodes, while the
other three are connected to exactly four nodes. Thus the total of
the nine weights must always be an even number, whichever numbers
are placed at the nodes.

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Now the smallest possible weight is $1+2=3$ while the largest
possible weight is $5+6=11$. As there are exactly nine edges, we
deduce that for the weights for each edge to be different they must
take the values $3, 4, 5, 6, 7, 8, 9, 10$ and $11$. However, the
total of these is $63$, an odd number, so the task is
impossible.

*This problem is taken from the UKMT Mathematical Challenges.*

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