On Sat, Apr 21, 2012 at 11:18 PM, Victor S. Miller <victorsmiller@gmail.com> wrote:
Let j be a prime >= k/2. If n has exactly j divisors then n=p^(j-1) for some prime p.
As I read the question, n can have more than j divisors, but it has exactly j divisors that are less than or equal to k.
If k>= 10 then 2^(k/2-1) > k so the statement is false for that j,k.
Victor
Sent from my iPhone
On Apr 21, 2012, at 16:16, David Wilson <davidwwilson@comcast.net> wrote:
For 1 <= j <= k, does there always exist n with exactly j divisors <= k?
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