(a) For each integer $k\ge 3$, find a positive integer $n$ that can be represented as the sum of exactly $k$ mutually distinct positive divisors of $n$.
(b) Suppose that $n$ can be expressed as the sum of exactly $k$ mutually distinct positive divisors of $n$ for some $k\ge 3$. Let $p$ be the smallest prime divisor of $n$. Show that \[\frac1p+\frac1{p+1}+\cdots+\frac{1}{p+k-1}\ge1.\]
(a) There is a known cute result that for $n\geq 3$ the factorial $n!$ can be written as the sum of $n$ distinct divisors of itself; see http://www.artofproblemsolving.com/Forum/viewtopic.php?f=151&t=373265&hilit=sum+of+distinct+divisors. We will thus take $n=k!$.
(a) for $ k=3 $ we have $ 6=1+2+3 $ and for $ k=4 $ we have $ 12=6+3+2+1 $.
if $ t $ can be written as the sum of $ k $ distinct divisors then $ 6t=3t+2t+t $ can be written as the sum of $ k+2 $ distinct divisors.
then we can make induction of step $ 2 $.
(b) we multiply the inequality by $ n $.
now is easy: the numbers obtained in $ LHS $ are greater or equal than any other $ k $ distinct divisors of $ n $.
now we now that we have $ k $ such dvisors that sum up $ n $ so the conclusion follows.
note: we used that the divisors in the sum of $ k $ terms are $ <n $ when $ k>1 $.