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Note that, $n×n!=(n+1-1)n!=(n+1)!-n!$ $S=\sum_{i=1}^{10}n×n!$

Telescoping $S$ with the help of above result,

we get $S=11!-1$ (How)

$\Rightarrow S+2=11!+1$ $\textbf {Clearly remainder is 1}$

$S = 1! + (2-1)2! + (3 - 1)3! + .................. + (11 -1)10!$

$S = 11! -1$

$S +2 = 11! +1$

Therefore remainder * 1

$S$ can be written as:

$\begin{aligned} S & = \sum_{n=1}^{10} n(n!) \\ & = \sum_{n=1}^{10}\big((n+1)n! - n!\big) \\ & = \sum_{n=1}^{10} (n+1)! - \sum_{n=1}^{10} n! \\ & = 11! - 1! \end{aligned}$

Therefore, $S+2 \equiv 11!-1!+2 \equiv \boxed 1 \text{ (mod 11)}$ .

We know that, $n\quad\times\quad n!=\quad(n+1-1)n!=(n+1)!-n!$ now let's see the given series

S=1!+2.2!+3.3!+........+10.10!

seeing the formula above we can say that,

S= 1! + (2+1)!-2!+(3+1)!-3!+.......+(10+1)!-10!

or, S=1!+3!-2!+4!-3!+5!-4!+6!-5!+7!-6!+8!-7!+9!-8!+10!-9!+11!-10!

S=1!-2!+3!-3!+4!-4!+........+10!-10!+11!

so, S=11!-2!+1! = 11!-2+1 = 11!-1

now adding 2 on both sides we get , S+2 = 11!+1

As 11!=1.2.3........11 , which is divisible by 11 , hence the remainder is 1