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

Nuber problem

  1. Apr 15, 2005 #1

    JamesU

    User Avatar
    Gold Member

    What's the next line in this problem? (don't look it up and be cheap.)
    1
    11
    21
    1211
    111221
    312211
    13112221
    1113213211
    31131211131221
    13211311123113112211
    11131221133112132113212221
    3113112221232112111312211312113211
    ????????????????????????????????????????

    Please put the answer in white, and go as many lines as you can. :smile:
     
  2. jcsd
  3. Apr 15, 2005 #2

    JamesU

    User Avatar
    Gold Member

    NE1 gonna try???? :bugeye:
     
  4. Apr 15, 2005 #3
    1
    11
    21
    1211
    111221
    312211
    13112221
    1113213211
    31131211131221
    13211311123113112211
    11131221133112132113212221
    3113112221232112111312211312113211
    1321132132111213122112311311222113111221131221111131221131211131231121113112221121321132132211331222113112221

    The line underneath explains the line above.. e.g. 123 = one 1, one 2, one 3 = 111213... etc..


    Arghh... gets tiring.... I probably made a mistake somewhere in there too... oh well, you get the jist.. :biggrin:
     
  5. Apr 15, 2005 #4

    JamesU

    User Avatar
    Gold Member

    It's not really hard, but what are all the extra 1's for in your first answer?
     
  6. Apr 16, 2005 #5
    I have NO idea :rofl: :biggrin:
     
  7. Apr 16, 2005 #6

    JamesU

    User Avatar
    Gold Member

    :rofl: :rofl: :rofl: :rofl:
    :rofl: :rofl: :rofl: :rofl:
    :rofl: :rofl: :rofl: :rofl:
    :rofl: :rofl: :rofl: :rofl:
    :rofl: :rofl: :rofl: :rofl:
    :rofl: :rofl: :rofl: :rofl: :rofl: :rofl: :rofl:

    :rofl: :rofl: :rofl: :rofl: :rofl:
    :rofl: o:) o:) :rofl: :rofl:
    :rofl: o:) o:) :rofl: :rofl:
    :rofl: o:) o:) :rofl: :rofl:
    :rofl: o:) o:) :rofl: :rofl:
    :rofl: :rofl: :rofl: :rofl: :rofl:

    :rofl: :rofl: :rofl: :rofl:
    :rofl: :rofl: :rofl: :rofl:
    :rofl: :rofl: :rofl: :rofl:
    :rofl: :rofl: :rofl: :rofl:
    :rofl: :rofl: :rofl: :rofl:
    :rofl: :rofl: :rofl: :rofl: :rofl: :rofl: :rofl:
     
  8. Apr 20, 2005 #7
    Is it...
    1321132132111213122112311311222113111221131221 :rofl:
     
    Last edited: Apr 20, 2005
  9. Apr 22, 2005 #8

    JamesU

    User Avatar
    Gold Member

    yes. :wink:
     
  10. Apr 22, 2005 #9
    Kinda fun, that one is.
    I had one of my friends on that for about a week one time, trying to solve it mathematically.

    Paden Roder
     
  11. Apr 25, 2005 #10

    JamesU

    User Avatar
    Gold Member

    I took some science class during the summer. Our teacher didn't have much to say, so he gave us riddles, that was one of 'em :rolleyes:
     
  12. Mar 22, 2009 #11
    1
    11
    21
    1211
    111221
    312211
    13112221
    1113213211
    31131211131221
    13211311123113112211
    11131221133112132113212221
    3113112221232112111312211312113211
    1321132132111213122112311311222113111221131221
    11131221131211131231121113112221121321132132211331222113112211
    311311222113111231131112132112311321322112111312211312111322212311322113212221
    132113213221133112132113311211131221121321131211132221123113112221131112311332111213211322211312113211
    11131221131211132221232112111312212321123113112221121113122113111231133221121321132132211331121321231231121113122113322113111221131221
     
    Last edited: Mar 22, 2009
  13. Mar 23, 2009 #12
    Its a really good brain teser. But is there anything special about the series other than that?
     
  14. Mar 23, 2009 #13
    Not really. Let's see...

    There's only 1 number that starts off with a 2-- all the rest start with 1 or 3. All of them end in a 1.

    Each digit in the series (left-aligning them) results in a 3-or-less length repeating sequence. So the 3rd digit (say) gets stuck in a loop starting at the 8th number in the series, repeating "1", "1", "2", such that the Nth number in the sequence's 3rd digit is predictable for N>=8. Each place follows a similar pattern with a loop length of 3 or 1. The 5th digit (for example) is always 1 starting at the 9th number in the sequence. And right-aligning the numbers results in 4-or-less length repeating sequences.

    For each number in the sequence, if there are A 1's, B 2's and C 3's, then A >= B >= C. The percentage occurrences of 1's, 2's, and 3's seem to be convergent to roughly 49.5%, 32%, and 18.5% respectively (looking down to about the 50th number in the sequence).

    There are probably some more things on that level of interesting, but nothing unexpected.

    DaveE
     
  15. May 9, 2009 #14
    Dave
    Thanks for the explaination. I couldn't see this page since long.
     
  16. May 19, 2009 #15
    Where does one start with this kinda stuff i'm a newbee
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Nuber problem
  1. A problem (Replies: 16)

  2. Communication problems (Replies: 10)

  3. The problem of time (Replies: 30)

  4. Explosive Problem (Replies: 6)

  5. Problems with Capitalism? (Replies: 142)

Loading...