Peter wrote: Find all positive integers $ n$ with the property that the set \[ \{n,n+1,n+2,n+3,n+4,n+5\}\] 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.

from piegonhole we know that if there are six numbers one of them must be divisible by 5. but we need product and so 2 numbers must be divisible by 5 and those numbers are $n,n+5$ so $n$ and $n+5$ are in the diferent subsets , we have 6 cases : 1° ${n,n+2,n+4}$ , and ${n+1,n+3,n+5}$ 2° ${n,n+3,n+4]}$ , and ${n+1,n+2,n+5}$ 3° ${n,n+1,,n+4}$ , ${n+2,n+3,n+5}$ 4° ${n,n+1,n+2}$ , ${n+3,n+4,n+5}$ 5 ${n,n+1,n+3}$ , ${n+2,n+4,n+5}$ 6 ${n,n+3,n+4}$ , ${n+1,n+2,n+5}$ from now just a litle work on inequalitles and none of these cases workt

Not so nice way (but better than the mod 5):

Nice way:

More generally, for any sequence of consecutive positive integers $n, n+1, \ldots, n+k$ there exists a prime $p$ such that $p$ divides $n+i$ for some $i$ with $0 \le i \le k$, but $p$ does not divide any of the other integers. This general statement implies that for any partition of a set of consecutive integers into at least two sets, there exists one set such that the product of the numbers in this set is different from any product of numbers in other sets. Proof. If $n > k+1$ we have Sylvester's Theorem that one element of our sequence is divisible by a prime $p > k+1$. None of the other elements can be divisible by $p$, since $p$ is larger than the length of the interval. For $n \le k+1$ we have a prime $p$ with $n \le \left \lceil(n + k)/2 \right\rceil < p \le n + k$ by Bertrand's Postulate and no other integer in the sequence can be divisible by $p$, since $2p > n + k$.