Chinese Mathematical Olympiad

Let $f(X)$ be the average of elements of the finite number set $X$.

First of all, make $n$ different primes $p_1,p_2,\dots ,p_n$ which are all bigger than $n$, and we prove that for any different nonempty subsets $A$, $B$ of the set $S_1=\{\prod (1/p_j):1\le j\le n\}$, $f(A)\ne f(B)$ always holds.

In fact, we can suppose that ${\prod }_{i=1}^n{p}_i\in A$ and ${\prod }_{i=1}^n{p}_i\notin B$ without loss of generality. Every element of $B$ can be divided by $p_1$, so $p_1\mid n!f(B)$. But $A$ has exactly one element which cannot be divided by $p_1$, so we find that $n!f(A)$ cannot be divided by $p_1$ (note that $p_1>n$), and therefore $n!f(A)\ne n!f(B)$, which follows $f(A)\ne f(B)$.

Second, let $S_2=\{n!x:x\in S_1\}$. Then $f(A)$ and $f(B)$ are different positive integers when $A,B$ are different nonempty subsets of $S_2$.

In fact, it is easy to see that there exist two sets $A_1,B_1$ which are different nonempty subsets of $S_1$, and $f(A)=n!f(A_1)$, $f(B)=n!f(B_1)$ holds. We get $f(A)\ne f(B)$ from $f(A_1)\ne f(B_1)$, and $f(A),f(B)$ are positive integers from $|A|,|B|\le n$ and their elements are all positive.

Then, let $K$ be the largest element of $S_2$. We prove that for every two distinct subsets $A,B$ of the set $S_3=\{K!x+1:x\in S_2\}$, $f(A)$ and $f(B)$ are coprime integers which are both larger than $1$.

In fact, it is easy to see that there exist two sets $A_1,B_1$ which are different nonempty subsets of $S_2$, and $f(A)=K!f(A_1)+1$, $f(B)=K!f(B_1)+1$ holds. Obviously, $f(A)$ and $f(B)$ are different integers which are both larger than $1$. If they have common divisors, let $p$ be a prime common divisor of them without loss of generality. Clearly, we have $p\mid (K!\cdot |f(A_1)-f(B_1)|)$. We get $1\le |f(A_1)-f(B_1)|\le K$ by $0<f(A_1),f(B_1)\le K$ and $f(A_1)\ne f(B_1)$, so $p\le K$, which follows $p\mid K!f(A_1)$, and then $p\mid 1$ — a contradiction.

Lastly, let $L$ be the largest element of $S_3$. We prove that for every two distinct nonempty subsets $A,B$ of the set $S_4=\{L!+x:x\in S_3\}$, $f(A)$ and $f(B)$ are two composites which share no common divisors.

In fact, it is easy to see that there exist two sets $A_1,B_1$ which are different nonempty subsets of $S_3$, and $f(A)=L!+f(A_1)$, $f(B)=L!+f(B_1)$ holds. Obviously, $f(A)$ and $f(B)$ are different integers which are both larger than $1$. Because of that, $L$ is the largest element of $S_3$, and we have $f(A_1)\mid L!$ and $f(A_1)\mid f(A)$. We find that $f(A)$ is composite by $f(A_1)<f(A)$. For a similar reason, $f(B)$ is composite too. If they have common divisors, let $p$ be a prime common divisor of them without loss of generality. It is obvious that $p\mid (L!\cdot |f(A_1)-f(B_1)|)$. We get $1\le |f(A_1)-f(B_1)|\le L$ by $0<f(A_1),f(B_1)\le L$ and $f(A_1)\ne f(B_1)$, so $p\le L$, which follows $p\mid f(A_1)$ and $p\mid f(B_1)$ — a contradiction of the fact that $f(A_1)$ and $f(B_1)$ are coprime. This completes the proof.