Find the set of $n\in\Bbb Z^+$ with $M=\{n,n+1,n+2,n+3,n+4,n+5\}$ partitionable into two sets 
Find the set of all positive integers $n$ with the property that the set $M=\{n, n + 1,n + 2,n + 3,n + 4,n + 5\}$ can be partitioned into two sets such that the product of the numbers in one set equals the product of the numbers in the other set.

If $n=1$ them $M=\{1,2,3,4,5,6\}$ and there is no such partition, so $n \ge 2$.
If the prime $p|n$ then either $p|2$ or $p|3$ or $p|5$ which means either $p=2$ or $p=3$ or $p=5$.
Suppose $n=2k$. Then $M=\{2k, 2k + 1,2k + 2,2k + 3,2k + 4,2k + 5\}$. I have no idea how to proceed.

Hint 1: Start by proving that each set of the partition must have 3 elements. To do this just argue that, for $n \geq 2$, the product of the smallest 4 elements is higher than the product of the largest 2. 
Hint 2: For all $n >0 $ we have
$$(n+3)(n+4)(n+1)>(n+5)(n+2)n$$
Hint 3 one of the sets of your partition must contain two of $n+3, n+4, n+5$. Now use 
$$(n+3)(n+4)(n+1)>(n+5)(n+2)n$$
to argue that the remaining number must be $n$.
Therefore, one of your set contains $n$ and two of $n+3, n+4, n+5$.
In each of the three cases, you get a quadratic equation.
