# product of the numbers in each subset is equal [duplicate]

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 ?

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## marked as duplicate by Ross Millikan, Jyrki Lahtonen♦, Jennifer Dylan, William, MJDAug 30 '12 at 16:08

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.

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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.

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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.

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That's not true if 5 divides $n$. – Gerry Myerson Jun 29 '12 at 12:22
+1 Thanks for finishing my answer. I had the feeling that the answer must look somewhat like yours but for some reason couldn't make it precise. – Simon Markett Jun 29 '12 at 13:47