Alice, Bob and his 1956-Triumph [closed]

Solution 1:

(it's essentially solution from @Vepir's comment, slightly modified to make proof simpler)

If $x \in B$, then some number from $[\frac{x}{20}, x]$ is in $A$ (we can't get $x$ by using numbers greater than $x$, or by using at most $20$ numbers smaller then $\frac{x}{20}$). So the idea is to use $20$ numbers from $B$ to make $20$ non-overlapping intervals for elements of $x$ (thus ensuring that each interval contains exactly one element from $A$), and then adding one more element to ensure all the numbers are from right border of corresponding intervals.

To do it, let $A = \{1, 10^2, 10^4, \ldots, 10^{38})$ and $B = A \cup \{1010\ldots1\}$. Let $A'$ be any list s.t. all numbers from $B$ can be sums of it. From the above argument, $A'$ contains a number from $[\frac{1}{20}; 1]$, from $[5, 100]$, from $[500, 10000]$, etc. As this intervals are non-overlapping, it contains exactly one number from each interval. If it contains any number that is not a right border of some of this intervals, then sum of even all elements of $A'$ is less than $1 + 100 + \ldots + 10^{38}$ - thus $B$ contains number that isn't sum of elements of $A'$. So $A$ contains only right borders of this intervals, so $A' = A$.