There are $n-1$ consecutive integers such that it is possible to divide them into two disjoint sets $A$ and $B$ such that the product of numbers in $A$ is congruent to product of numbers in $B$ modulo $n$.
We have to verify the above statement for $n=13$ and $n=19$.
For $n=13$ I have proved such sets exists.
$A = \{1,2,3,4,5,6\}$ and $B = \{7,8,9,10,11,12\}$
Then we have the required condition.
I think for $n=19$ it is not true. How to disprove that?
Please help.
Assume $x$ is product of all numbers in $A$ and $y$ is product of all numbers in $B$.
We want $x \equiv y \pmod {19}$. As $19$ is prime, it's equivalent to $x y^{-1} \equiv 1 \pmod {19}$. Which is equivalent to $(xy)y^{-2} \equiv 1 \pmod {19}$. But $xy$ is just $18!$. By Wilson theorem, $18! \equiv (-1) \pmod {19}$. So, we have $y^2 \equiv -1 \pmod{19}$.
But by Euler's criterion, $-1$ is quadratic residue modulo $19$ iff $(-1)^{\frac{19 - 1}{2}} = (-1)^9$ is equal to $1$. As this is not true, there is no $y$ s.t. $y^2 \equiv -1 \pmod{19}$, and so no such $A$ and $B$.