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Math Help - LCM(a,b,c) and abc

  1. #1
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    LCM(a,b,c) and abc

    Theorem: LCM(a,b,c)·F=abc
    F=GCD(a,b)·GCD(ab/GCD(a,b),c)

    Proof:
    1) LCM(a,b)·GCD(a,b)=ab
    2) LCM(a,b,c)=LCM(LCM(a,b),c) or see below
    3) Let H=gcd(a,b), then
    4) LCM(a,b)=ab/H
    4) LCM(a,b,c)=LCM(ab/H,c)
    5) LCM(a,b,c)=[abc/H]/GCD(ab/H,c)
    6) LCM(a,b,c)·H·GCD(ab/H,c)=abc,

    EXAMPLE
    LCM(4,6,15)x2xGCD(2x6,15)=4x6x15
    LCM(4,6,15)x2x3=4x6x15
    LCM(4,6,15)=60
    You get same answer for any order of a,b,c
    ---------------------------------------------------------
    LCM(a,b,c)=LCM(LCM(a,b),c), by prime factorization which is easy to see for specific cases but awkard to type in the general case, or:
    7) Let Q=LCM(LCM(a,b),c)
    8) a|LCM(a,b), LCM(a,b)|Q, → a|Q
    9) b|LCM(a,b), LCM(a,b)|Q, → b|Q
    10) c|Q

    So Q is a common multiple of (a,b,c). Now you have to show Q divides every common multiple of (a,b,c). Then Q =LCM(a,b,c)

    CM(a,b,c) =ra=sb=tc
    ra=sb =CM(a,b)
    CM(a,b,c)=CM(a,b)=tc
    Q|CM(a,b), Q|c → Q|CM(a,b,c)

    The beauty of the proof is its easy to type.
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  2. #2
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    Re: LCM(a,b,c) and abc

    Quote Originally Posted by Hartlw View Post
    Theorem: LCM(a,b,c)·F=abc
    F=GCD(a,b)·GCD(ab/GCD(a,b),c)
    You can't prove this, it is not true.
    Counter-example: a= 6, b= 8, c= 12. LCM(6, 8, 12)= 2 and 6(8)(12)= 576 so F= 576/2= 488.
    But GCD(6, 8)= 2 and (6)(8)/GCD(6, 8)= 48/2= 24. GCD(24, 12)= 12. There product is 24, not 488.

    Proof:
    1) LCM(a,b)·GCD(a,b)=ab
    2) LCM(a,b,c)=LCM(LCM(a,b),c) or see below
    3) Let H=gcd(a,b), then
    4) LCM(a,b)=ab/H
    4) LCM(a,b,c)=LCM(ab/H,c)
    5) LCM(a,b,c)=[abc/H]/GCD(ab/H,c)
    6) LCM(a,b,c)·H·GCD(ab/H,c)=abc,

    EXAMPLE
    LCM(4,6,15)x2xGCD(2x6,15)=4x6x15
    LCM(4,6,15)x2x3=4x6x15
    LCM(4,6,15)=60
    You get same answer for any order of a,b,c
    ---------------------------------------------------------
    LCM(a,b,c)=LCM(LCM(a,b),c), by prime factorization which is easy to see for specific cases but awkard to type in the general case, or:
    7) Let Q=LCM(LCM(a,b),c)
    8) a|LCM(a,b), LCM(a,b)|Q, → a|Q
    9) b|LCM(a,b), LCM(a,b)|Q, → b|Q
    10) c|Q

    So Q is a common multiple of (a,b,c). Now you have to show Q divides every common multiple of (a,b,c). Then Q =LCM(a,b,c)

    CM(a,b,c) =ra=sb=tc
    ra=sb =CM(a,b)
    CM(a,b,c)=CM(a,b)=tc
    Q|CM(a,b), Q|c → Q|CM(a,b,c)

    The beauty of the proof is its easy to type.
    Thanks from topsquark
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  3. #3
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    Re: LCM(a,b,c) and abc

    The proof is correct. I tried various a,b,c. There may be a typo, or you made a mistake. Will clear it up tomorrow morning. I do tend to screw up details of typing. Thanks for checking
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  4. #4
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    Re: LCM(a,b,c) and abc

    Quote Originally Posted by HallsofIvy View Post
    You can't prove this, it is not true.
    Counter-example: a= 6, b= 8, c= 12.
    Au contraire mon ami.

    If (a,b,c) = 6,8,12
    GCD(6.8) = 2
    F = 2xGCD(6x8/2,12) = 2x12
    LCM(6,8,12)x2x12 = 6x8x12
    LCM(6,8,12) = 24

    It’s really a simple derivation. All you need is LCM(a,b,c)=LCM(LCM(a,b),c), LCM(a,b)xGCD(a,b)=ab, and some simple algebra. It’s probably easier to do yourself than follow mine.

    The formula can be recursive:
    LCM(a,b,c,d) = LCM(LCM(a,b,c),d), to give
    LCM(a,b,c,d)xF = abcd,
    and F depends only on GCD’s of two numbers, which are easy to work with, but formula may be a little messy to write out.
    So in general,
    LCM(a,b,c,d,….)xF=abcd….
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  5. #5
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    Re: LCM(a,b,c) and abc

    HallsofIvy, Also note the order of a,b,c doesn’t matter:

    Theorem: LCM(a,b,c)•F=abc
    F=GCD(a,b)•GCD(ab/GCD(a,b),c)

    If (a,b,c) = (12,6,8)
    GCD(12,6) = 6
    F = 6xGCD(12x6/6,8) = 6x4
    LCM(12,6,8)x6x4 = 12x6x8
    LCM(12,6,8) = 24
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