Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

A Circle And Adjacent Number Puzzle

  1. Aug 25, 2007 #1
    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.
    Last edited: Aug 25, 2007
  2. jcsd
  3. Aug 25, 2007 #2
    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

    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

  4. Aug 25, 2007 #3
    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.
    Last edited: Aug 25, 2007
  5. Aug 25, 2007 #4
    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....

  6. Aug 25, 2007 #5
    Yes, hence the subordinate clause "For the neighboring condition to hold". Placing three points as such dosen't satisfy the condition.
  7. Aug 25, 2007 #6

    Please, read again:

    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...)

  8. Aug 25, 2007 #7
    I see, I had understood something else, sorry.
    Last edited: Aug 26, 2007
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook