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Math Help - Series

  1. #1
    Newbie darknight's Avatar
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    Series

    The numbers 1,2,3........1998 are written in natural order.
    Numbers in odd places are stricken off to obtain a new sequence.
    2,4,6,........
    Then, numbers in odd places of this sequence are stricken off.
    which will obtain: 4,8,......
    This process continues until only one term 'a' is left. Then a = ?????

    I started by,
    2,4,6..........,1996, 1998
    Using Arithmetic Progression:
    1998 = 2 + (n-1)2
    1996/2 = n-1
    n = 998 + 1
    n = 999

    Therefore, in the next step 1998 will be stricken off
    New sequence = 4,8,12......, 1992, 1996
    Again, by A.P.
    1996 = 4 + (n-1)4
    1992/4 = n-1
    n = 499
    So, new sequence is:
    8,16, 24..... 1984, 1992

    So far:
    2,4,6..........,1996, 1998
    4,8,12......, 1992, 1996
    8,16, 24..... 1984, 1992

    I noticed that the differences were increasing multiples of 2.
    1996 + 2 = 1998
    1992 + 4 = 1996
    And tried to deduce a, using arithmetic and geometric progressions but couldn't figure out how.

    Since the number of terms in the sequence was exponentially decreasing, I used brute force.
    I went on till the starting number of the sequence became 512:
    and the sequences looked like this:
    256, 512, 768,.... 1376
    512, 1024... 1120

    And completing the last iteration I got 'a' as 1024.

    Is this correct? And secondly, isn't their a simpler way to solve this?
    The calculation was long and tedious, and I worked hard, not smart.
    Finally, this is a 2 mark question (out of a 60 mark test), so I'm sure there is a simpler way.
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  2. #2
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by darknight View Post
    Is this correct? And secondly, isn't their a simpler way to solve this?
    The calculation was long and tedious, and I worked hard, not smart.
    Finally, this is a 2 mark question (out of a 60 mark test), so I'm sure there is a simpler way.
    It's correct. The answer will always be the greatest power of two less than or equal to the upper limit. I came to this conclusion from reviewing what you wrote. The reason is that it's impossible to eliminate any power of 2 from the list until it's the first element in the list.

    A little more explanation: The first time you went through, you only removed odd numbers. The second time you went through, you only removed numbers that were divisible by 2 but not 4. Then you removed numbers divisible by 4 but not 8. Etc.
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  3. #3
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    1,2,3,4,5,6,7,8,9,10,11,12.......1998.....
    2,4,6,8,10,12,14,16,18,20...................
    4,8,12,16,20,24,28,32,36....................
    8................................................. ...
    2^n............................................... .

    You can think this way:
    the first term actually will be the last term of the sequence after cutting the sequence n-1 times .
    So you only need to find 2^n<=1998
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  4. #4
    Newbie darknight's Avatar
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    Thanks for the insight.
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  5. #5
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    Quote Originally Posted by darknight View Post
    Thanks for the insight.
    No problem.

    Just check my last post (I have changed it), it is much neater solution.
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