Oh dear 11!

Consider k(k!)=(k+1)!-k!, the sum equals to 101!-1. Since 11 is a factor of 101!, the sum=(-1) (mod 11) = 10 (mod 11)

Since k(k!) is divisible by 11 if $k\ge11$ ,

$\sum_{k=1}^{100}k(k!)$ $\sum_{k=1}^{10}k(k!)\mbox{ (mod 11)}$ $\equiv 10\times10! + 9\times9! + 8\times8! + \cdots + 3\times3! + 2\times2! + 1\times1!\mbox{ (mod 11)}$ $\equiv 10\mbox{ (mod 11)}$

$k(k!)=(k+1-1)(k!)=(k+1)!-(k!)$

therefore, the given sum reduces to:

$2!-1!+3!-2!+4!-3!+.....+101!-100!=101!-1$

And since $101!$ is divisible by $11$ , the remainder is $-1$ . Remainders are positive by convention, so we find the positive remainder counterpart, which is $10$

To offer a different solution... As 11| 11! + 12! +...+ 101! $\sum_{k=1}^{100} (k+1)!\equiv(2! + 3! +...+ 10!)\equiv10mod(11)$ (I used a calculator to find the remainder of the final sum)

dis cld be solvd simply see frm 11 11! to lst term der wl be no remainder if we divide by 11. nw for summ k k! for 1 to 10 cab ve solvd as k*k! =k+1!-k! .nw alternate trms are cancld nd we r left wid 11!-1! whch is 11 n-1 frm nd =s to
11(n-1)-10 nd wen divided by 11 gvs 10 as remaindr