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  1. #1
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    Prove:

    For any integer n>1, prove that there is exactly one power of 2 having exactly n digits with leading digit 1
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  2. #2
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    Quote Originally Posted by newtoinequality View Post
    For any integer n>1, prove that there is exactly one power of 2 having exactly n digits with leading digit 1
    Hi

    It can be done by induction
    The property is true for n=2 (16 = 2^4)

    Suppose that it is true for n. This means that there exists p integer such that 10^{n-1} \leq 2^p < 2 \cdot 10^{n-1}

    You have to demonstrate that there exists q integer such that 10^{n} \leq 2^q < 2 \cdot 10^{n}
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  3. #3
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    Quote Originally Posted by running-gag View Post
    Hi

    It can be done by induction
    The property is true for n=2 (16 = 2^4)

    Suppose that it is true for n. This means that there exists p integer such that 10^{n-1} \leq 2^p < 2 \cdot 10^{n-1}

    You have to demonstrate that there exists q integer such that 10^{n} \leq 2^q < 2 \cdot 10^{n}
    You mean 10^n\leq 2^{n+1}< 2\cdot 10^{n}, don't you?
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  4. #4
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    Quote Originally Posted by newtoinequality View Post
    For any integer n>1, prove that there is exactly one power of 2 having exactly n digits with leading digit 1
    First, show existence. If the statement were false, then there would exist m,\,n such that 2^m<10^{n-1} and 2^{m+1}\ge2\cdot10^{n-1}. But the former gives 2^{m+1}<2\cdot10^{n-1} which is a contradiction of the latter.

    Hence for each n there must be m such that 10^{n-1}\le2^m<2\cdot10^{n-1}, i.e.

    2^m\ <\ 2\cdot10^{n-1}\quad\ldots\ \fbox1

    2^m\ \ge\ 10^{n-1}\quad\ldots\ \fbox2

    To prove uniqueness, observe that 2^{m-1}<10^{n-1} (from \fbox1) and 2^{m+1}\ge2\cdot10^{n-1} (from \fbox2). The result follows since if m'\ne m then either 2^{m'}\le2^{m-1} or 2^{m'}\ge2^{m+1} and so 2^{m'} cannot be between 10^{n-1} and 2\cdot10^{n-1}.
    Last edited by TheAbstractionist; July 23rd 2009 at 01:08 PM.
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  5. #5
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    Quote Originally Posted by HallsofIvy View Post
    You mean 10^n\leq 2^{n+1}< 2\cdot 10^{n}, don't you?
    No

    For instance 10^3 \leq 2^{10} < 2 \cdot 10^3
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