product of the numbers in each subset is equal 
Possible Duplicate:
product of six consecutive integers being a perfect square 

Find  all  positive  integers $n$ such  that 
the set  {$n, n + 1, n + 2, n + 3, n + 4, n + 5$} can be partitioned into two subsets 
so that the product of the numbers in each subset is equal.
One possible way I think to solve this is to consider this set $\mod 5$ and check for the partitions for which the product modulo 5 comes equal and then solve for those partitions to get the integer values of $n$. Is this a fine approach? If yes, are there other methods to solve this easily as I think my procedure is time consuming ?  
 A: Revised after seeing there were $6$ rather than $5$ numbers to partition.
As Simon Markett remarks, $n$ has to be divisible by $5$, and no other prime factors than $2$ and $3$ occur in the numbers $n+1,n+2,n+3,n+4$. Two of them are even and of the remaining two only one can be divisible by $3$ which leaves the final one without prime factors; as these numbers are${}>1$, this is impossible.
A: The condition implies that the product of the six consecutive numbers is a square. But it's known that the product of two or more cosecutive numbers can't be a square. For references, and discussion of the 6-case, see this earlier question. 
A: If one of the numbers is divisble by a prime, then at least one other number has to be divisible by the same prime. Otherwise you will never be able to partition them into two sets with equal product.
But this can only happen for primes $\leq 5$, so all $6$ numbers just have the factors $2$, $3$ and $5$. In fact at least one and at most two of them will be divisble by five so you have to have $5|n$ (and not $5^2|n$). 
Moreover at least one of the numbers is divisible by $6=2\cdot 3$, so neither of its neighbours is divisible by either $2$ or $3$.
It feels like there is no solution.
