1

This problem is meant to be proved using contradiction and I've admittedly been stuck on it for some time. I'd really appreciate a point in the right direction, because it honestly feels like I've been getting nowhere.

This is all I've done:

Assume $x$ denotes the set of nine consecutive integers, and $x_1$ and $x_2$ denote the two partitions. We would have: $$\prod_{a\ \in \ x_1} a= \prod_{a\ \in \ x_2} a$$ $\implies$ $\prod_{a\ \in \ x_1} a$ and $ \prod_{a\ \in \ x_2} a$ are both even numbers, since both sets $x_1$ and $x_2$ would require to contain strictly odd numbers to have a product that is odd.

$\implies$ $x_1$ has at least two elements, one odd and one even.

$\implies$ $x_2$ has at most seven elements.

And that's where I hit a dead end. I thought perhaps I could represent the set $x$ as $$x := \{ (a_0, \ (a_0 +1), \ (a_0 +2), \ \dots , \ (a_0 +8) \}$$ where $a_0$ is any integer, then multiply things out and show in each case one side is less than the other, but then I noticed that'd be an extremely tedious route and that some of the more ugly expansions deserve some sort of proof by induction themselves to show whether or not they actually hold.

Please don't supply a proof that relies too heavily on a Number Theory-approach, as I don't have a background in that area therefore shouldn't rely on Theorems there to prove this.

Andrew Tawfeek
  • 2,654

1 Answers1

1

Suppose we partitioned the set into two sets $A,B$ such that $ A \cup B= \{a, a+1, a+2 ... a+8 \} $ such that the product is equal: $$ \prod_{n\in A}n=\prod_{n\in B}n $$ Here multiplying both sides by $\prod_{n\in B}n $ leads to $$ \left(\prod_{n\in A}n\right)\left(\prod_{n\in B}n\right)=\left( \prod_{n\in B}n \right)^2 \ \ \ \Longrightarrow \ \ \ a(a+1)(a+2)...(a+8)=\left( \prod_{n\in B}n \right)^2 $$ But by Erdos-Selfridge theorem the product of two or more consecutive integers is never a power, I.E. we have contradiction

Michele Caselli
  • 714