## A Circle And Adjacent Number Puzzle

Analytically determine if it is possible to arrange the numbers 0,1,2,3,4,5,6,7,8,9 (not necessarily in this order) in a circle such that the difference between any two adjacent numbers is 3, 4 or 5.

Note: Each of the ten numbers must occur exactly once.
 PhysOrg.com science news on PhysOrg.com >> City-life changes blackbird personalities, study shows>> Origins of 'The Hoff' crab revealed (w/ Video)>> Older males make better fathers: Mature male beetles work harder, care less about female infidelity
 Consider the loop from 9 to 9 (a sequence encompassing the other numbers). Well, immediately after the first 9 and before the last 9 you can have only 4,5 or 6. And after (and before) the number 1, you can have only 4,5 or 6. So, without loss of generality, lets consider the 1 is the third number in the sequence. Then we have 9,[4,5,6],1,[4,5,6],X,X,X,X,[4,5,6],9 Now, lets consider the numbers 0 and 8. Theirs neighbors have to be [3,4,5] but 4 and 5 are already allocated. So, the 3 has to be between 0 and 8, and this group of 3 numbes (0,3,8) has to be starting or ending the interval X,X,X,X,X. So, the remaining 2 positions are bounded by 0 or 8, which lefts no space for the 2 and 7. So, the proposed arrangement is not possible!
 Rogerio, a simpler, more straightforward, explanation exists: Arranging three points on a circle (here 3,4 and 5) creates three arcs each defined as the space between two points. For the neighboring condition to hold, an arc can only harbor two points. Hence, only six other points can be fitted onto the circle; we need seven.

## A Circle And Adjacent Number Puzzle

 Quote by Werg22 Rogerio, a simpler, more straightforward, explanation exists: Arranging three points on a circle (here 3,4 and 5) creates three arcs each defined as the space between two points. For the neighboring condition to hold, an arc can only harbor two points. Hence, only six other points can be fitted onto the circle; we need seven.
How do you prove that? I don't think it is so "straightforward" as you mean...

In fact, you could place three points between the 4 and the 5: ...4,1,6,9,5....

 Yes, hence the subordinate clause "For the neighboring condition to hold". Placing three points as such dosen't satisfy the condition.

 Quote by Werg22 Yes, hence the subordinate clause "For the neighboring condition to hold". Placing three points as such dosen't satisfy the condition.
?!?!?!