Can 44 be a factor of $(n-3)(n-2)(n-1)n(n+1)(n+2)(n+3)$ where n is a natural number greater than 3?

Question

Can 44 be a factor of $(n-3)(n-2)(n-1)n(n+1)(n+2)(n+3)$ where n is a natural number greater than 3 ?

When $n =4 $,

$(n-3)(n-2)(n-1)n(n+1)(n+2)(n+3)=1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7$

Answer 1

Certainly, $4$ is a factor. Let $n=11m+k\ (0\le k\le 10)$. If $4\le k\le 7$ then $11$ is not a factor, else it is.

Answer 2

Any natural number of the form $$y = x\text{ mod 11 }$$ where $x = 4,5,6,7$ will not be a solution. All others will.

For example, $$15 = 4 \text{ mod } 11$$ $$12\cdot{13}\cdot{14}\cdot{15}\cdot{16}\cdot{17}\cdot{18}$$ is not a solution but $$19 = 8 \text{ mod 11 }$$ $$16\cdot{17}\cdot{18}\cdot{19}\cdot{20}\cdot{21}\cdot{22}$$ is a solution