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Math Help - Prove n is divisible by 2^m

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    Prove n is divisible by 2^m

    Prove that if n is a positive integer such that the integer which is made up from the last m digits of n in its decimal representation is divisible by 2^m then n is divisible by 2^m
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by mandy123 View Post
    Prove that if n is a positive integer such that the integer which is made up from the last m digits of n in its decimal representation is divisible by 2^m then n is divisible by 2^m
    Decompose n into a number k_1 made from its last m digits and the number
    made by its remaining digits:

    n=k_2 10^{m} +k_1

    Now we are told that 2^m|k_1, so to complete this problem therefore
    it is sufficient to show that 2^m|k_210^m

    RonL
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    Forum Admin topsquark's Avatar
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    I know what the problem is asking and I don't dispute the result, but I don't like how its phrased. For example, 4|100, but "00" is divisible by anything. It probably should be mentioned as a special case for the problem to be stated correctly. (Yes, I'm in a picky mood this morning.)

    -Dan
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