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Alc, blc, then (ab)lc

  1. Dec 2, 2008 #1
    1. The problem statement, all variables and given/known data
    alc, blc, then (ab)lc
    I'm trying to explain why this is true

    2. Relevant equations

    3. The attempt at a solution
    a is a factor of c an b is a factor of c.
    ab is a factor of c is true because a will still be a factor of c and b will still be a factor of c.
  2. jcsd
  3. Dec 2, 2008 #2
    I don't know where I'm going wrong
  4. Dec 2, 2008 #3


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    Other people do know where you are going wrong. And they told you why the first time you posted it. Why don't you read the first post? Stop it.
  5. Dec 2, 2008 #4
    Ok, I think i figured it out. It's not true because we can have 6l6 and 2l6, but not 12l6.
  6. Dec 2, 2008 #5
    Sorry, but I never posted this before.
  7. Dec 3, 2008 #6


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  8. Dec 3, 2008 #7
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