21 Apr
2012
21 Apr
'12
4:19 p.m.
Let j be a prime >= k/2. If n has exactly j divisors then n=p^(j-1) for some prime p. 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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