RE: [PrimeNumbers] Re: An introduction.
- From: firstname.lastname@example.org [mailto:email@example.com]
On Behalf Of Mark Underwood
> multiples of the prime p. Regarding whether the converse is true, Iwould suspect so.
> That is, given a row whose interior numbers are all multiples of thatrow number, is the
> row number necessarily prime? The formula is n!/(k!*(n-k)!) , (kfrom 0 to n) , and if
> n=p*v, (p prime, v>1) I'm pretty sure it can be shown that when k is amultiple of p then
> n!/(k!*(n-k)!) will not have a factor of p, and hence will not be amultiple 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
p which divides this term is p^(m-1). This shows n does not divide this
Indeed and old gem, but as Mark says, nice to rediscover!