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Using all the numbers 4, 5, 6, ... 10 exactly once, fill each empty circle with a number so that if two numbers are in circles that are joined by a line then their difference (in absolute value) is at least 3.

Ten circles, seven empty.  Others contain 1, 2 and 3

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  • $\begingroup$ Is my solution missing anything, or leaving out anything? $\endgroup$ Jul 18, 2023 at 19:24
  • $\begingroup$ @newQOpenWid Your answer could be better if you added more explanation to your deductive reasoning. $\endgroup$ Jul 18, 2023 at 19:29
  • $\begingroup$ thanks for the feedback, added some logic explanation $\endgroup$ Jul 18, 2023 at 19:39

1 Answer 1

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This is the solution I found:

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The top rectangle must have $4$, $7$ and $10$, as all the numbers are connected to each other and have to be 3 apart. But $7$ has to go in the top circle because of the requirement in the bottom: The node under $7$ would have to be either $4$ or $10$, and that would be impossible.

So then we can deduce the rest of the graph: If $10$ went in the left node, then the node under it would have to be either $6$ or $5$. Therefore the bottom-left corner node has to be $9$ or $8$. Similarly for the right node, $4$, the node under that would have to be $9$, since if it were $8$, the node under that would not have any number to go with. So then the node under that has to be $6$.

Now, the bottom-left corner node must be 8 since 9 is used up. But that is impossible since that violates the requirement. Henceforth, contradiction, meaning $4$ has to go in the left node and $10$ has to go in the right node.

We can then apply the same argument, but this time we do not get a contradiction, but rather a winning answer.

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