TITLE: Find $x \in \mathbb Z^+$ such that $x\sum _{i=1}^{x} \frac 1{i!} \in \mathbb Z$
QUESTION [6 upvotes]: Find $x \in \mathbb Z^+$ such that 
$$n = x\cdot \left ( \frac 1{1!} + \frac 1{2!} + \cdots +\frac 1{x!} \right )  \in \mathbb Z$$

It is easy to show $x = 1$ is a solution, 
But I don't know what to do to find all the $x$, please give me a hint.
Thank you !

REPLY [4 votes]: Hint:
$$
x\cdot \left ( \frac 1{1!} + \frac 1{2!} + \cdots +\frac 1{x!} \right )=\frac{x!+\frac{x!}{2}+...+x+1}{(x-1)!}
$$
and for $x > 3$, the numerator is never divisible by $3$, while the denominator...
Hence, the only solutions are...