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RE: [PrimeNumbers] Re: An introduction.

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  • Chris Caldwell
    From: primenumbers@yahoogroups.com [mailto:primenumbers@yahoogroups.com] On Behalf Of Mark Underwood ... would suspect so. ... row number, is the ... from 0 to
    Message 1 of 3 , Dec 4, 2008
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      From: primenumbers@yahoogroups.com [mailto:primenumbers@yahoogroups.com]
      On Behalf Of Mark Underwood
      > multiples of the prime p. Regarding whether the converse is true, I
      would suspect so.
      > That is, given a row whose interior numbers are all multiples of that
      row number, is the
      > row number necessarily prime? The formula is n!/(k!*(n-k)!) , (k
      from 0 to n) , and if
      > n=p*v, (p prime, v>1) I'm pretty sure it can be shown that when k is a
      multiple of p then
      > n!/(k!*(n-k)!) will not have a factor of p, and hence will not be a
      multiple of n.

      Suppose n is not prime and let p be a prime divisor of it (as above).
      Now let n = (p^m)*v
      where p does not divide v (e.g., is n=12, p=2, then m=2 and v=3.).

      Now consider n choose p, this is

      [ (p^m*v) * (p^m*v-1) *...* (p^m*v-p+1) ] / [ p * (p-1) * ... * 1 ]

      Note p only divides the first term of the top and bottom, so the power
      of
      p which divides this term is p^(m-1). This shows n does not divide this

      coefficient.

      Indeed and old gem, but as Mark says, nice to rediscover!

      CC
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