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A proof of divisibility problem

  1. Feb 5, 2013 #1
    If a divides b, and a divides b+c then a divides 3c.

    How do I go forward with this?

    This is what I've done so far:

    suppose a|b and a|b+c

    then b = an for some integer n
    and c = am for some integer m

    ∴ b+c = an+am
    = a(n+m)
    = ak for some integer k

    but I don't feel like this is getting me anywhere
     
  2. jcsd
  3. Feb 5, 2013 #2
    Alright, I got it

    If a|3c then a is a multiple of 3

    then b is also a multiple of 3, as a divides it

    therefore a|b+c = b/a + c/a

    and thus a|c

    therefore a|3c
     
  4. Feb 7, 2013 #3
    why doesn't 'a' divide c and not 3. Honestly the teacher put 3c in the problem to confuse you. He could have made it 13c. That wouldn't mean 'a' divides 13 since it can be shown that 'a' divides c. P.S you have shown that c = am so what does 3c equal?
     
  5. Feb 7, 2013 #4
    we start out by assuming that a|3c is correct,
    therefore a has to be a multiple of 3, if it divides 3c
     
  6. Feb 7, 2013 #5
    No 'a' can be 7, 'c' can be 14 and 'b' can be 35. 3c would then be 42, a divides 3c because a divides c which is am. 3c would be 3am. a divides 3am because the factor a is in the product 3am. there is nothing in the problem that would suggest that 'a' divides 3.
     
    Last edited: Feb 7, 2013
  7. Feb 7, 2013 #6
    but it divides it into a multiple of 3. That's my point
     
  8. Feb 7, 2013 #7
    That doesn't mean that 3 divides 7,14 and 21. Thus neither a, b or c need be multiples of 3which is contrary to what you said.
     
  9. Feb 7, 2013 #8
    hmm.. alright, point taken..
    so how do I go about this then?
     
  10. Feb 7, 2013 #9
    All you need to do is show that c = am, so that 3c = 3am = a multiple of a.
     
    Last edited: Feb 7, 2013
  11. Feb 8, 2013 #10

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    3c= 3(b+ c)- 3b.

    Really what you are proving is that "if a divides b+ c and a divides b, then a divides c." The "3" is irrelevant.
     
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