<s>[INST] 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.
 [/INST] Partition $M$ into $A$ and $B.$ For brevity let  $A^*=\prod_{x\in A}x$ and $B^*=\prod_{y\in B}y.$ Assume that $A^*=B^*$.
We may assume $n+5\in B.$
Observe that $5|n$  otherwise exactly one member of $M$ is divisible by $5,$ implying that $5$ divides exactly one of $A^*, B^*.$ So let  $M=\{5m,5m+1,5m+2,5m+3,5m+4,5m+5\}.$
Now let $p$ be any prime divisor of $m+1.$
(1).If $p=5$ then $5^2 | 5(m+1),$ so $5^2 | B^*, $ but then the only member of $A$ that is divisible by $5$ is $5m,$ which is not divisible by $5^2.$   Therefore  $p\ne 5.$ 
(2). So $p| (m+1)$, and $p|B^*$ and $p\not |\;  5m,$ so $p$ divides at least one member of $\{5m+j:1\leq j\leq 4\} $ (Otherwise $p\not |\; A^*.$) Now $p|(5m +5)$ and $p|(5m+j)$ for some $j\in \{1,2,3,4\},$ implying $p|(5m+5)-(5m+j)=5-j$ for some $j\in \{4,3,2,1\}.$ So the only possible prime divisors of $m+1$ are $2$ and $3.$ 
(3). If $3|(m+1)$ then the only other member of $M$ that is divisible by $3$ is $5m+2,$ so $5m+2\in A$ and $5m+5\in B.$ And $3^2$ cannot divide either $5m+2$ or $5m+5,$ otherwise exactly one of $A^*,B^*$ is divisible by $3^2.$ Therefore if $3|(m+1)$ then $3^2\not |\; (m+1).$
(4). If $2|(m+1)$ then $5m+1,5m+3, 5m+5$ are even while $5m, 5m+2,5m+5$ are odd. Now if $2^4|(m+1)$ then $2^4|B^*, $ but also $2^4\not |\; (5m+3)(5m+1)$, implying $2^4\not |\; A^*.$   Therefore if $2|(m+1)$ then $2^4\not |\; (m+1).$ 
(5).Therefore $m+1\in \{2^a3^b:  a\leq 3\land b\leq 1\}$ \ $\{1\}=\{2,4,8, 3, 6, 9, 12\}.$ So $n+5=5m+5\in\{10,20,40,15,30,45,60\}.$ But for each of these potential values for $n+5,$ there is a prime $q\geq 7$ that belongs to  $M,$ and does not divide any other member of $M, $ so $q$ divides exactly one of $A^*,B^*.$
So $A^*\ne B^*$ and there are no solutions.
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