The problem has 2 questions:
a)Prove that for any $n\geq 3$ there exists $x_1<x_2<...<x_n$ such that $\frac1{x_1}+\frac1{x_2}+...+\frac1{x_n}=1$
b)Prove that for any $n\geq3$ there exist $n$ natural numbers such that their least common multiple can be written as the sum of $n$ of their divisors(distinct).
For a) I chose: $$x_1=2, x_2=4, ... x_{n-2}=2^{n-2}, x_{n-1}=2^{n-2}+1, x_n=2^{n-2}(2^{n-2}+1)$$ which meets the 2 requirements.
Unfortunately I can't seem to be able to crack b). I tried using the same numbers from a)(their least common multiple being $x_n$) but I can't write it as a sum of $n$ of its divisors. In fact I think it might not be possible. I also played around with other powers of 2 but with no success.
Thanks for your help!