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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.

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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