Determine all positive integers $n$ such that $$n!\left(1+\dfrac{1}{2}+\ldots+\dfrac{1}{n}+\dfrac{1}{n!}\right)$$ is divisible by $n$.
Proposed by Alessandro Ventullo, Milan, Italy
Solution:
Clearly, $n=1$ satisfies the condition. Let $n>1$. Then, $$n!\left(1+\dfrac{1}{2}+\ldots+\dfrac{1}{n}+\dfrac{1}{n!}\right)=n!\left(1+\dfrac{1}{2}+\ldots+\dfrac{1}{n-1}\right)+(n-1)!+1.$$ It's easy to see that $$n!\left(1+\dfrac{1}{2}+\ldots+\dfrac{1}{n-1}\right) \equiv 0 \pmod{n}.$$ Furthermore, by Wilson's Theorem, $(n-1)!+1 \equiv 0 \pmod{n}$ if and only if $n$ is prime, so $n$ must be a prime number.
In conclusion, $n=1$ or $n=p$, where $p$ is a prime number.
