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Very easy question

  1. Jan 19, 2007 #1
    I forgot how to prove the following, can anyone jog my memory:

    Let m,n be natural numbers s.t. gcd(m,n) = 1 and n|k and m|k for some natural number k.

    Then nm|k.

    I know I'll kick myself when I find out. Thanks.
  2. jcsd
  3. Jan 19, 2007 #2
    well what do n|k and m|k represent and what does it imply about k? think in terms of mathematical symbols and not words. Apply the gcd constriction and then you should be done.
  4. Jan 19, 2007 #3
    This also falls to the fundamental theorem of arithmetic nicely.
  5. Jan 19, 2007 #4
    Oh I see it now!

    You just decompose m and n into their unique prime factors and note that no prime factor of m can be a prime factor of n and vice versa.

    So if am = k = bn then each prime factor of n, say, divides a (with aprropriate multiplicity) and hence n divides a giving the result.

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