A peer of mine showed me earlier today this problem, taken from a 7th grade math contest :
Let $A=\{1,2,3,\ldots,n\}$; (where $n \geq 5$) prove that $A$ can be divided into two disjoint subsets such that the sum of the elements in the first subset is equal to the product of the elements in the second subset.
This has been puzzling me for 15 minutes already, but I'm sure there's a simple, straight-forward way to do it since it's a 7th grade problem, albeit I can't see it.
Can anyone shed some wisdom here ?
Best Answer
Going by the assumption that for an odd $n$ the product of $1, (n-1), (n-a)$ is equal to the sum of the rest of the numbers, we have $$ 1(n-1)(n-a) = \frac{n(n+1)}{2} - 1 -(n-1) - (n-a) $$
solving this gives $a=\frac{n+1}{2}$
Now for odd $n \ge 5$, we have
$$n-a = \frac{n-1}{2} \ge 2$$ Therefore the numbers $1, n-1, n-a$ are distinct.
A similar assumption for even $n$ with $1, n, (n-a)$ gives $$ 1*n(n-a) = \frac{n(n+1)}{2} - 1 -n - (n-a) $$ and the solution is $a=\frac{n+2}{2}$
And for even $n \ge 6$ we have $$n-a = \frac{n-2}{2} \ge 2$$ And therefore the numbers $1, n, n-a$ are distinct.