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

  • $\begingroup$ 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


This is the solution I found:

enter image description here

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.