Oh dear 11!

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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