75th NMO Selection Tests

Assume $a_1\le a_2\le \cdots \le a_n$. If we choose the subset $B^{\prime }=\{1,2,\dots ,2^{n-2}\}$, then among the sums $s_1,s_2,\dots ,s_{2^{n-2}}$, the values $a_1,a_2,\dots ,a_t$ will appear for some $t\ge 1$. Clearly, $s_1=s_{\{1\}}=a_1$, and if $a_i$ appears as a sum, then $a_{i-1}$ must have appeared before it. Indeed, consider:

a) if $a_{i-1}=a_i$, then the appearance is obvious.

b) if $a_{i-1}<a_i$, then since $s_{\{i-1\}}=a_{i-1}<a_i$ and $a_i$ appears as some $s_j$, the index corresponding to $a_{i-1}$ is less than $j$, hence $a_{i-1}$ appeared earlier.

If $t\le n-2$, then at most $2^{n-2}-1$ sums can appear, corresponding to non-empty subsets of $\{1,2,\dots ,n-2\}$. Hence, we deduce that $t\ge n-1$. Therefore, the first two sums correspond to the numbers $a_1$ and $a_2$. Then, inductively, after determining the numbers $a_1,a_2,\dots ,a_k$, we compute all the subset sums of $\{a_1,a_2,\dots ,a_k\}$, remove them from the list of known sums, and the smallest remaining sum must be $a_{k+1}$. So, we can determine the numbers $a_1,a_2,\dots ,a_{n-1}$. If we also include the index $2^n-1$, we obtain $s_{2^n-1}$, which is the total sum of all numbers. Then we can compute $a_n$, and thus all numbers.