n=1 satisfies the j=1 case for all k. n=2 satisfies the j=2 case for all k. The first slightly interesting case is j=3. For k=3, n=6 is the only possible solution, but for all larger k, n=4 works. For j=4: if j=4 or 5, n=12. If j=6, n=6 is, I believe, forced. And n=6 works for all larger k. In general, we can say that for j=k, k! works (and there are smaller solutions). And if n is the smallest number with exactly j divisors, it works for all k>=n. But in the gap, things can probably get tricky. On 4/21/12, Mike Stay <metaweta@gmail.com> wrote:
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?
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
-- Mike Stay - metaweta@gmail.com http://www.cs.auckland.ac.nz/~mike http://reperiendi.wordpress.com
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun