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