The sum of elements of a set equals the product of another one .(Math contest) 
We say that a natural number $n\ge 2$ is SP (i don’t what does that mean), if we can split the set $$S=\{1,2,3,...,n\}$$ into two nonempty subsets, (They don’t share any common element) such that the sum the first subset equals the product of the second one. Example: $n=7$ Is SP because $2+4+5+7=1
\cdot 3 \cdot 6$

*

*prove that $n=4$ is not SP


*check whether or not $6$ and $7$ are SP


*Prove that all $n\ge 7$ are SP

This is the problem number $3$ (which is the last one) in a math Olympiad in which i participated yesterday.
I didn’t do any question from these $3$ Ones (i mean the questions of the third problem), i thought that the best way to do the first question is to just guess and check cases, but i realized that there is a lot of cases the check so this method is not good, and then i’ve tried to find a general way but i didn’t get anything.
 A: 1: The SP-value has an upper bound $\sum_{k=1}^n k = \frac{n(n+1)}{2}$. Therefore for $n=4$ we get $|SP|<10$ and the product contains either $3$ or $4$. This reduces the  product values to $|SP|\in\{3,4,6,8\}$ - just 4 cases to check.
2,3: Using the example $2+4+5+7=6\cdot 3\cdot 1\Rightarrow \sum_{k=1}^7 k = 6\cdot 3\cdot 1 + 6+3+1$ we get a good starting point for a general formula. For odd numbers $2n+1$ we can use $$(2n+1)(n+1) = 2n\cdot n \cdot 1 + 2n +n+1$$ and for even $2n$:
$$n(2n+1) = 2n\cdot (n-1) \cdot 1 + 2n +(n-1)+1 .$$
A: Slightly different way of doing this:
$$n=8: \quad 8 \times 3 \times 1 = \left(\sum_{i=1}^8 i\right)-8-3-1$$
$$n=9:  \quad 8 \times 4 \times 1 = \left(\sum_{i=1}^8 i\right)-3-1 = \left(\sum_{i=1}^8 i\right)+9-8-4-1 = \left(\sum_{i=1}^9 i\right) -8-4-1.$$
We now find a way of partitioning.
Assume the following for an even integer $n$:

(IH): $$n \times \left(\frac{n}{2}-1\right) \times 1 = \left(\sum_{i=1}^n i\right)-n-\left(\frac{n}{2}-1\right)-1.$$

Then
$$n \times \left(\frac{n}{2}\right) \times 1 = n+\left(\sum_{i=1}^n i\right)-n-\left(\frac{n}{2}-1\right)-1$$
$$=n+1+\left(\sum_{i=1}^n i\right)-n-\frac{n}{2}-1 = \left(\sum_{i=1}^{n+1} i\right)-n - \frac{n}{2} - 1
.$$
So $n+1$ is SP.
And
$$(n+2) \times \frac{n}{2} \times 1 = 2n+\left(\sum_{i=1}^n i\right)-n-\left(\frac{n}{2}-1\right)-1
=(n+1) + \left(\sum_{i=1}^n i\right)-\frac{n}{2}-1$$
$$=\left(\sum_{i=1}^{n+1} i\right)- \frac{n}{2}-1 = \left(\sum_{i=1}^{n+2} i\right)-(n+2) - \frac{n}{2}-1.$$
So $n+2$ is SP as well, and (IH) holds when $n$ is replaced with $n+2$.
