A Circle And Adjacent Number Puzzle

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The discussion centers on the challenge of arranging the numbers 0 through 9 in a circle such that the difference between any two adjacent numbers is either 3, 4, or 5. Initial analysis reveals that certain numbers, such as 1, 4, 5, and 6, have restricted placements due to their adjacency requirements. Specifically, if 1 is positioned third in the sequence, it limits the placement of surrounding numbers. Further exploration shows that the numbers 0 and 8 must be adjacent to 3, 4, or 5, leading to a conflict in fitting all numbers within the constraints. Ultimately, the conclusion drawn is that such an arrangement is impossible, as the necessary conditions cannot be met without violating the adjacency rules. The discussion also touches on the geometric interpretation of arranging points on a circle, emphasizing that only six points can fit under the given conditions, while seven are needed, reinforcing the impossibility of the arrangement.
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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.
 
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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, let's 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, let's 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!

:smile:
 
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.
 
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Werg22 said:
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...

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

Please, read again:

...the difference between any two adjacent numbers is 3, 4 or 5.
THIS is the only neighboring condition.
And the difference between any two adjacent numbers from the sequence "4,1,6,9,5" IS 3, 4 or 5!

(BTW: the "difference between any two adjacent numbers" is always a non negative number...)

:smile:
 
I see, I had understood something else, sorry.
 
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