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Math Help - Breaking a stick of lenght of power of 2

  1. #1
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    Breaking a stick of lenght of power of 2

    hi, if there is a stick of lenght which of the form of power of 2 , then what is the minimum number of times one must break the stick to get a lenght , if the stick can only be broken into half each time.
    E.g if the stick is of lenght 8, then to get a length of 5 we must break it a minimum 3 times .
    Is there any general formula to it ?
    Thanks.
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  2. #2
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    Oh this is so my thing, for I, in the least, "like" (i.e. LOVE) the binary system.
    Anyway, consider 5 in the binary system - 5=101. Note that it's last digit is one. That also means that the highest power of 2 that divides it is 1 (i.e. 2 to 0).
    Now let's take another number e.g. 44.
    It can be written as 44=32+8+4=101100. Note the two zeroes from the right. It is the same as saying that the number is divisible by 4 (i.e. the second power of 2).
    So if understood you correctly, the problem you posed for n=44, would be something like this:
    (64)=(32,32)=(32,16,16)=(32,16,8,8)=(32,16,8,4,4) - 4 breakings
    And now you can construct 44 from these numbers...
    So if got it right your solution in a generic formula would be:
    x=\lceil \log_{2} n \rceil-d, where d is the power of the highest divisor of the number n.
    So I guess that's it?

    EDIT: Ah, I made a mistake by claiming you could write 44 with (32,16,8,8)... upon better inspection, I see the formula uses the ceil function rather than the floor but now I'm positive that it's correct.
    Last edited by Croat; June 8th 2011 at 11:50 AM. Reason: LATEX, just got a small thing wrong...
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