Question 75
Let $X = \{x_0, \dots, x_{n-1}\}$ be an $n$-element set of real numbers such that $0 \lt |x_0| \le \dots \le |x_{n-1}|$. Prove that the sums of elements of all subsets of $X$ are $2^n$ consecutive members of an arithmetic sequence in some order if and only if
$$
|x_0| : \dots : |x_{n-1}| = 2^0 : \dots : 2^{n-1}.
$$
$$
|x_0| : \dots : |x_{n-1}| = 2^0 : \dots : 2^{n-1}.
$$