China Girls' Mathematical Olympiad

The answer of the problem is the set of all positive real numbers less than $2$. We consider two cases.

Case I We assume that $0<a<2$. Then there is a positive $n$ such that $2^{n-1}>a^n$. We define $A_n=\{m\mid m\text{ is a multiple of }{2}^{n-1}\}$ and $$A_i=\{2^{i-1}m\mid m\text{ is an odd integer}\},$$ for $1\le i\le n-1$. Then $A_1,A_2,\dots ,A_n$ is a partition of the set of positive integers satisfying the conditions of the problem.

Case II We assume that $a\ge 2$. We claim that no such partition exists. To prove by contradiction, we assume on the contrary that $A_1,A_2,\dots ,A_n$ is a partition satisfying the conditions of the problem. Let $N=\{1,2,\dots ,2^n\}$. For every $i$ with $1\le i\le n$, let $B_i=A_i\cap N$. We assume that $B_i=\{b_1,b_2,\dots ,b_m\}$ with $b_1<b_2<\cdots <b_m$. We have $$2^n>b_m-b_1=(b_m-b_{m-1})+(b_{m-1}-b_{m-2})+\cdots +(b_2-b_1)\ge (m-1)2^i,$$ implying that $m-1<2^{n-i}$, or $m\le 2^{n-i}$. Since $A_1,A_2,\dots ,A_n$ is a partition, $B_i\cap B_j=\varnothing$ for $1\le i<j\le n$ and $N=B_1\cup B_2\cup \cdots \cup B_n$. It follows that $$2^n=|N|=|B_1|+|B_2|+\cdots +|B_n|\le \sum _{i=1}^n{2}^{n-i}=2^n-1,$$ which is impossible. Hence our assumption was wrong and such a partition does not exist for every positive integer $n$.