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Maths Sequence Problem

  • 22-04-2011 3:36pm
    #1
    Closed Accounts Posts: 84 ✭✭


    Ok, so my friend posed me this question the other day and for the life of me I can't figure it out. So I thought I might ask you guys to see would you have any ideas on the answer(s).

    First number: Unknown
    Second number: 15
    Third number: 16
    Fourth number: 21
    Fifth number: 23
    Sixth number: 31
    Seventh number: 111
    Eight number: Unknown
    Ninth number: Unknown

    There can be more numbers preceeding the first but you don't need to work them out.

    I wonder can any of you guys work out the three answers because I can't see any pattern at all.


Comments

  • Registered Users, Registered Users 2 Posts: 123 ✭✭gaeilgeboy


    Qutie the unusual sequence, not something I'd have ever worked out myself. I checked OEIS, and it gave that the [LATEX]n^{\text{th}}[/LATEX] term in this sequence:
    [LATEX]
    [/LATEX]
    [LATEX]{10, 11, 12, 13, 14, 15, 16, 21, 23, 31, 111, 1101, 1111111111111}
    [/LATEX]
    is [LATEX]13[/LATEX]
    written in base [LATEX]13-n[/LATEX], for [LATEX]n = 0, 1, \dots, 12[/LATEX].


  • Closed Accounts Posts: 6,081 ✭✭✭LeixlipRed


    Haha, what are the chances anyone would ever work that out by themselves? Interesting though!


  • Closed Accounts Posts: 84 ✭✭resistantdoor


    gaeilgeboy wrote: »
    Qutie the unusual sequence, not something I'd have ever worked out myself. I checked OEIS, and it gave that the [LATEX]n^{\text{th}}[/LATEX] term in this sequence:
    [LATEX]
    [/LATEX][LATEX]{10, 11, 12, 13, 14, 15, 16, 21, 23, 31, 111, 1101, 1111111111111}
    [/LATEX]
    is [LATEX]13[/LATEX] written in base [LATEX]13-n[/LATEX], for [LATEX]n = 0, 1, \dots, 12[/LATEX].

    Wow. Ok, thanks for the reply, but could you offer an insight into what the answer means and how the answer was derived?


  • Registered Users, Registered Users 2 Posts: 2,149 ✭✭✭ZorbaTehZ


    The basic unit we unit to count in is 10, i.e. the decimal system. Say you can write 147 as 1x10^2 + 4x10^1 + 7x10^0 and likewise 13 is 1x10^1 + 3x10^0. So you could use a different number as the "base" viz.:

    13 in base 13 is 1x13^1 + 0x13^0 = 10 (b13)
    13 in base 12 is 1x12^1 + 1x12^0 = 11 (b12)
    ...
    13 in base 3 is 1x3^2 + 1x3^1 + 1x3^0 = 111 (b3)
    ...
    and so on.


  • Registered Users, Registered Users 2 Posts: 2,651 ✭✭✭ShowMeTheCash


    Using base conversion i.e. Base 16 hex is common place in programming.
    The benefit really being notation and space. But if you are familiar with it, it is not too difficult to work out. If not familiar with it then understandably it might seem very difficult.


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