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Math Help - [SOLVED] Brainy 3

  1. #1
    Member u2_wa's Avatar
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    Arrow [SOLVED] Brainy 3

    Another interesting question.

    If you start counting from the thumb of your right hand, on which finger or if thumb, will you have 2009.

    An example is also attached to show you how to count.

    Attached Thumbnails Attached Thumbnails [SOLVED] Brainy 3-hand.jpg  
    Last edited by u2_wa; March 22nd 2009 at 02:31 AM.
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  2. #2
    Newbie dashed's Avatar
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    Im guessing the 4th (ring) finger?

    EDIT: nevermind, i just saw the diagram
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  3. #3
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    Count the first few numbers on each finger.

    Thumb: 1,9,17,...

    Index: 2,8,10,16,18,...

    Middle: 3,7,11,15,19,...

    Ring: 4,6,12,14,20,...

    Pinky: 5,13,21,...

    The sequences for the thumb, middle, and pinky fingers form an arithmetic progression, each given by 8n-7, 4n-1, and 6n-1, respectively. Since n has to be an integer, we test 2009 for each sequence. The pinky sequence satisfies that condition, therefore you will count 2009 on the pinky.

    If the question was asking to find on what finger will you count 2010 initially, then the only choice would be is to finger numbers close 2010 that satisfies the sequences for the thumb or pinky, and then count from there. But this is not applicable if you were to choose a number that satisfies the middle sequence, since you can't tell in what direction is it going.
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  4. #4
    Member u2_wa's Avatar
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    Thanks to all who tried!!
    The answer is thumb.
    Now let me prove.

    The equation of pinky that I've calculated is  5+8(n-1)
    5+8(n-1)=2009, n=251.5


    n should be an integral number so let n=251
    5+8(250)=2005 =2005

    Now count from 2005 i.e on pinky to left to 2009 which lands on thumb!!

    Quote Originally Posted by Chop Suey View Post
    Count the first few numbers on each finger.

    Pinky: 5,13,21,...

    The sequences for the thumb, middle, and pinky fingers form an arithmetic progression, each given by 8n-7, 4n-1, and \color{red}6n-1\color{black}, respectively. Since n has to be an integer, we test 2009 for each sequence. The pinky sequence satisfies that condition, therefore you will count 2009 on the pinky.
    I don't think so this equation 6n-1 is of sequence (Pinky: 5,13,21,...)
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  5. #5
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    Hehe sorry about that, this is what happens when you don't revise your work. But our approaches are the same.
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  6. #6
    Member u2_wa's Avatar
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    Quote Originally Posted by Chop Suey View Post
    Hehe sorry about that, this is what happens when you don't revise your work. But our approaches are the same.
    Yeh I agree!!!!
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