Find $x \in \mathbb Z^+$ such that $x\sum _{i=1}^{x} \frac 1{i!} \in \mathbb Z$

Question

Find $x \in \mathbb Z^+$ such that $$n = x\cdot \left ( \frac 1{1!} + \frac 1{2!} + \cdots +\frac 1{x!} \right ) \in \mathbb Z$$

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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 !

Answer 1

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...

Answer 2

$$\begin{align*}2\cdot\sum_{i=1}^2\frac1{i!}&=2\left(1+\frac12\right)=3\in\Bbb Z\\ 3\cdot\sum_{i=1}^3\frac1{i!}&=3\left(1+\frac12+\frac16\right)=3+\frac32+\frac12=5\in\Bbb Z\end{align*}$$