The product of all differences of the possible couples of six given positive integers is divisible by 960. How can I show that the the product of all differences of the possible couples of six given positive integers is divisible by $960$?
$$x_1≥x_2≥x_3≥x_4≥x_5≥x_6$$
$$960\mid (x_1-x_2 )(x_1-x_3 )(x_1-x_4 )(x_1-x_5 )(x_1-x_6 )(x_2-x_3 )(x_2-x_4 )(x_2-x_5 )(x_2-x_6 )(x_3-x_4 )(x_3-x_5 )(x_3-x_6 )(x_4-x_5 )(x_4-x_6 )(x_5-x_6)$$
 A: By the pigeonhole principle, at least two of the original numbers are equivalent modulo 3 and two are equivalent modulo 5. Thus the product of differences will be divisible by $3\times 5=15$.
Similarly, the mod 2 class values may be split 6:0, 5:1, 4:2 or 3:3. In these cases the difference product will have 15, 10, 7 or 6 powers of two included. Thus the difference product will be divisible by at least $2^6= 64$. 
$64\times 15 =960 $ as required.
(We could make a stronger claim; the product of differences should be divisible by $8640$ because of multiple matching mod 3 values.) 

edit to add - We could make the strongest possible claim by applying the pigeonhole principle to mod 4 numbers and thus gather an extra two powers of $2$, meaning all such numbers are divisible by $34560$ - and observe that this is also equal to the result, $34560=5!4!3!2!1!$, when the original numbers are a consecutive sequence giving minimum non-zero differences.
A: If $k$ are odd (or even) we have $k(k-1)/2$ combinations each yielding a power of $2$. This gives us the options $(6:0)=15, (5:1)=10, (4:2)=6+1=7$ and $(3:3)=3+3=6$. So we know $X$ is divisible by $2^6$. With six variables, we have at least $3^3$ from the pigeonhole principle, and similarly $5$ is a divisor.
