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Find the remainder when $\sum_{k=1}^{100} k(k!)$ is divided by 11.

Details:-

$\bullet$ $k!$ stands for factorial of $k$, that is $k!=k\times (k-1)\times (k-2) \times ... \times 2 \times 1$

This is a part of the set 11≡ awesome (mod remainders)

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