(k-10)!=77(k-11)! решите уравнение

[latex](k - 10)! = 77(k - 11)!\\\\ n! = 1\cdot 2 \cdot 3 \cdot ... \cdot (n - 1) \cdot n, \ 0! = 1; \ n! \ne 0, \ n \in \mathbb{N}; \ k \geq 11 \\\\ (k - 10)! = 77(k - 11)! \ | \ : \ (k - 11)!\\\\ \frac{(k - 10)!}{(k - 11)!} = 77\\\\ \frac{1 \cdot 2 \cdot (k - 11) \cdot (k - 10)}{1 \cdot 2 \cdot (k - 11)} = 77\\\\ k - 10 = 77, \ \boxed{k = 87}[/latex]