The set $\{1,2,3,\ldots,n\}$, where $n \geq 5$, can be divided into two subsets so that the sum of the first is equal to the product of the second 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 ?
 A: The sum of all the numbers is
$$
\sum_{k=1}^nk=\frac{n(n+1)}2\tag{1}
$$
We will find $a$ and $b$ so that
$$
\overbrace{\left(\sum_{k=1}^nk\right)-a-b-1}^{\text{sum of all but $a$, $b$, and $1$}}=\overbrace{\vphantom{\left(\sum_1^n\right)}1ab}^{\text{product}}\tag{2}
$$
That is, we need to find $a$ and $b$ so that
$$
\begin{align}
\frac{n(n+1)}2
&=\overbrace{ab}^{\text{product}}+\overbrace{a+b+1}^{\text{sum}}\\
&=(a+1)(b+1)\tag{3}
\end{align}
$$
and $(3)$ can be solved using
$$
(a,b)=\left\{\begin{array}{}
\left(\frac{n-1}2,n-1\right)&\text{if $n$ is odd}\\
\left(n,\frac{n-2}2\right)&\text{if $n$ is even}\\
\end{array}\right.\tag{4}
$$
Thus, the product of $1$, $a$, and $b$ is the sum of the rest of the numbers from $1$ to $n$.
If $n\ge5$, then $a,b\ge2$ so that they don't interfere with $1$.

Examples
With $n=5$, $a=\frac{5-1}2=2$ and $b=5-1=4$ so $1\cdot2\cdot4=3+5$
With $n=6$, $a=6$ and $b=\frac{6-2}2=2$ so $1\cdot2\cdot6=3+4+5$
With $n=7$, $a=\frac{7-1}2=3$ and $b=7-1=6$ so $1\cdot3\cdot6=2+4+5+7$
With $n=8$, $a=8$ and $b=\frac{8-2}2=3$ so $1\cdot3\cdot8=2+4+5+6+7$
A: 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.
A: For any odd $n$ and $n\geq5$ we can always find the item before the last one, and the item before the center item, and the first item. Then move those 3 items to the product set. 
For example: $n=5$, move $1,2,4$. If: $n=7$, move $1,3,6$. 
For any even $n$ and $n\geq6$ we can always find the last item, and the item just before the 2 center items, and the first item. Then move those 3 items to the product set. That is the solution.
For example: $n=6$, move $1,2,6$. If: $n=8$, move $1,3,8$. 
for $n=3$, move 3. It seems that this is a special case.
To proof, we have the following relationships:
For: $n=odd$
Left: 
$${1\over2}n(n+1)-(n-1)-{(n-1)\over2}-1={1\over2}(n-1)^2$$
Right: 
$$(n-1){(n-1)\over2}={1\over2}(n-1)^2$$
For: $n=even$
Left: 
$${1\over2}n(n+1)-n-({n\over2}-1)-1={1\over2}n(n-2)$$
Right: 
$$n({n\over2}-1)={1\over2}n(n-2)$$
Note:
If we want to just give an answer to the question, then the above is OK. But in general, we need to know (1) How many items to move? (2) Which ones to move.  We cannot assume there must be 3 items, and 1 and (n-1) must be within the choices. Further we do not know if it is the only solution. We may need to proof.
