Iranian Mathematical Olympiad

a) Let $a_1\le a_2\le \cdots \le a_n$ be the stolen numbers. Since all the partial sums are positive, all the stolen numbers must be positive. Obviously $a_1$ is the smallest number among the partial sums. Suppose that numbers $a_1,a_2,\dots ,a_i$ have been determined. Omit all the partial sums of $a_1,a_2,\dots ,a_i$ from the partial sums of $a_1,a_2,\dots ,a_n$. The smallest number among the remaining numbers, must be $a_{i+1}$, so all the numbers will be determined uniquely.

b) Suppose that $a_1\le \cdots \le a_k<0\le a_{k+1}\le \cdots \le a_n$ are the stolen numbers and $s_1\le s_2\le \cdots \le s_{2^n-1}$ are the partial sums. Note that if $s_1>0$, the problem is already solved in part (a). Therefore, we can assume that $s_1<0$. We have $$(1+x^{a_1})(1+x^{a_2})\cdots (1+x^{a_n})=1+x^{s_1}+x^{s_2}+\cdots +x^{s_{2^n-1}}.$$ Obviously, $s_1$ is the sum of negative numbers $a_1,a_2,\dots ,a_k$. Multiplying the above equation by $x^{-s_1}$ implies $$(1+x^{-a_1})\cdots (1+x^{-a_k})(1+x^{a_{k+1}})\cdots (1+x^{a_n})=x^{-s_1}+x^{s_1-s_1}+x^{s_2-s_1}+\cdots +x^{s_{2^n-1}-s_1}.$$ Hence we can say that we have the partial sums of stolen numbers $|a_1|,|a_2|,\dots ,|a_n|$, and by part a, it is possible to find them uniquely. Finally, we must show that $x^{s_1}$ can be uniquely written as a product of $x^{a_i}$'s. If there were two ways to do this, we would obtain a partial sum of $a_i$'s equal to 0, which leads to a contradiction.

c) The partial sums of two sets $\{1,2,-3\}$ and $\{-1,-2,3\}$ are the same, so adding 1389 0's to these sets will not change the partial sums.