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Math Help - interger problem

  1. #1
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    interger problem

    Every positive integer can be expressed in a unique way as the sum of powers of 2. Express 2008 as the sum of powers of 2.
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  2. #2
    Junior Member
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    Oct 2008
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    Just calculate the logarithms with base 2 and round down:
    log(2008)/log(2) = 10.97... ==> 10
    log(2008-2^10)/log(2) = 9.94... ==> 9
    log(2008-2^10-2^9)/log(2) = 8.88... ==> 8
    log(2008-2^10-2^9-2^8)/log(2) = 7...
    log(2008-2^10-2^9-2^8-2^7)/log(2) = 6...
    log(2008-2^10-2^9-2^8-2^7-2^6)/log(2) = 4...
    log(2008-2^10-2^9-2^8-2^7-2^6-2^4)/log(2) = 3 exactly

    So we have 2^10+2^9+2^8+2^7+2^6+2^4+2^3=2008.
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