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The question was stimulated by this one. Here it comes:

When you look at the sum $\sum\limits_{k=1}^N k!$ for $N\geq 10$, you'll always find $3$ and $11$ among the prime factors, due to the fact that $ \sum\limits_{k=1}^{10}k!=3^2\times 11\times 40787. $ Increasing $N$ will give rise to factors $3$ resp. $11$.

Are $3$ and $11$ the only common prime factors in $\sum\limits_{k=1}^N k!$ for $N\geq 10$?

I think, one has to show, that $\sum\limits_{k=1}^{N}k!$ has a factor of $N+1$, because the upcoming sum will always share the $N+1$ factor as well. This happens for $ \underbrace{1!+2!}_{\color{blue}{3}}+\color{blue}{3}! \text{ and } \underbrace{1!+2!+\cdots+10!}_{3^2\times \color{red}{11}\times 40787}+\color{red}{11}! $

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    @Chickenmancer yes common for all sums like$3$and 11 for N>102018-02-26

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