Find the remainder of the factorials

$\begin{aligned} P & = \sum_{n=1}^{10} n\times n! \\ & = \sum_{n=1}^{10} \left((n+1)\times n! - n!\right) \\ & = \sum_{n=1}^{10} \left((n+1)! - n!\right) \\ & = 11! - 1! \\ \implies P+2 & = 11! - 1 + 2 \\ & \equiv (11!+1) \text{ (mod 11)} \\ & \equiv \boxed{1} \text{ (mod 11)} \end{aligned}$

$P+2=2+(1×1!)+(2×2!)....+(10×10!)$ $=1+1+(1×1!)+(2×2!).......+(10×10!)$ $=1+1+1+4+(3×3!)+(4×4!)...+(10×10!)$ $=1+6+(3×3!)+(4×4!)......+(10×10!)$ $=1+3!+(3×3!)+(4×4!)....+(10×10!)$ $=1+3!(1+3)+(4×4!)....+(10×10!)$ $=1+4!+(4×4!)....+(10×10!)$ $=1+4!(1+4)+(5×5!)...+(10×10!)$ $=1+5!+(5×5!)...+(10×10!)$ Similarly $=1+10!+(10×10!)$ $=1+11!$ Then P+2 divided by 11 $=(1+11!)/11$ $=1/11 +11!/11$ Thus remainder will be 1