$\displaystyle 1!+2!+\dots +1000!\equiv x \ \mbox{(mod 11)}$

When $\displaystyle k\geq 11, k!\equiv 0 \ \mbox{(mod 11)}$

Since we are adding up to 11!, can this be done without $\displaystyle \sum_{i=1}^{10}k!$ individual?

$\displaystyle 1!+2!+\dots +10!\equiv x \ \mbox{(mod 11)}$