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.
 A: 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.
A: 
Proposition.  Let $p$ be a prime natural number such that $p\equiv3\pmod{4}$.  Then, for any integer $k$, the set $\{k+1,k+2,k+3,\ldots,k+p-1\}$ cannot be partitioned into two sets $A$ and $B$ such that $$\prod_{a\in A}\,a\equiv \prod_{b\in B}\,b\pmod{p}\,.$$

If $k$ is not divisible by $p$, then exactly one element amongst $k+1,k+2,\ldots,k+p-1$ is divisible by $p$.  Thus, any partition of $S:=\{k+1,k+2,\ldots,k+p-1\}$ into two subsets $A$ and $B$ satisfies $$\prod\,X\equiv 0\pmod{p}\text{ and }\prod\,Y\not\equiv0\pmod{p}\,,$$
with $\{X,Y\}=\{A,B\}$.  
If $k$ is divisible by $p$, then $$\prod\,S\equiv(p-1)!\equiv-1\pmod{p}$$ by Wilson's Theorem.  If $S$ had a partition into two subsets $A$ and $B$ such that $\prod\,A\equiv \prod\,B\pmod{p}$, then $$x^2\equiv\left(\prod\,A\right)\,\left(\prod\,B\right)=\prod\,S\equiv-1\pmod{p}\,,$$
where $x:=\prod\,A$, but this is absurd as $-1$ is not a quadratic residue modulo $p$.

For this problem, take $p:=7$ and set $k:=n-1$.  As a trivial consequence of the proposition above, we conclude that there does not exist such an integer $n$.
A: Hint, given
$$n \equiv 0 \pmod{n}$$
$$n+1 \equiv 1 \pmod{n}$$
$$n+2 \equiv 2 \pmod{n}$$
$$n+3 \equiv 3 \pmod{n}$$
$$n+4 \equiv 4 \pmod{n}$$
$$n+5 \equiv 5 \pmod{n}$$
if there exists such a partition, $Q\cap S=\varnothing$ and $Q\cup S=\left \{ 0,1,2,3,4,5 \right \}$ 
$$\prod_{q \in Q} \left ( n+q \right ) = \prod_{s \in S} \left ( n+s \right ) \Rightarrow \prod_{q \in Q} \left ( n+q \right ) \equiv \prod_{s \in S} \left ( n+s \right ) \pmod{n}$$
which is the same as
$$\prod_{q \in Q} q \equiv \prod_{s \in S} s \pmod{n}$$
And $0$ will be on one side, let's suppose the right one
$$\prod_{q \in Q} q \equiv 0 \pmod{n}$$
which has no solutions for $n > 120$. This leaves us with just a handful of cases to analyse.
A: 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.
