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Increasing counterclockwise number of integers

  1. Mar 9, 2014 #1
    1. The problem statement, all variables and given/known data
    In the image below we start with the integer 1 marked in yellow. We will fill the rest of the table in counterclockwise manner with integers to infinity. What will be the sum of the numbers right above and below the number 2008?

    attachment.php?attachmentid=67448&stc=1&d=1394398270.png

    2. Relevant equations
    -


    3. The attempt at a solution
    I think it has something to do with positive/ negative integers.
     

    Attached Files:

  2. jcsd
  3. Mar 9, 2014 #2
    This reminds me of a project Euler problem. The key is to find patterns in the numbers, locate where that number would be, and then do a little grunt work to find the numbers above and below it. For example, the most apparent pattern to me is what happens with the numbers to the SE of 1. Draw out a 11x11 (or bigger) grid and see what you can find.
     
  4. Mar 10, 2014 #3
    That idea also came to my mind. There is some pattern here that I at least haven't observed yet. I've tried to map it out on a larger grid and still wasn't able to find anything.

    Maybe one additional piece of information, this question was presented as part of a 45 min/ 30 question test to assess your abilities for a business game... To me it seems that there must be something obvious about this that I'm missing.
     
  5. Mar 10, 2014 #4

    Curious3141

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    Here's a strategy that might work. First find out which arm of the 'spiral' 2008 will occur in. Then figure out roughly where in the spiral (at either end or somewhere in the middle?) of that arm it will occur. Finally try to work out the relationship between numbers in the middle of one of that sort of arm with the adjacent ones.

    To get you started on the first part, notice that 2, 10, 26, ... always seem to occur at the extreme ends of the 'rightward' spiral. Can you see the pattern that governs these numbers? How long are these rightward spiral arms?
     
  6. Mar 11, 2014 #5

    lurflurf

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    Answer the question for (2n+1)^2-k
    k<n
    then let n=22,k=17

    If general k seems hard do k=1,2,..
    until you see the pattern
     
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