If the sum of the digits of n is equal to the sum of the digits of 5n, then prove that 9|n. Let $n\in\mathbb{N}$.
So far I have: If the sum of the digits of $n$ is $k$, then $n = 9m + k$, where $m$ element of an integer (not sure why). Now consider $5n-n$.
Help?

You know that $n=9i + k$ and $5n = 9j+ k$. So $4n = 9(j-i)$ is divisible by $9$. But since $\gcd(4,9)=1$, we must have that $n$ is divisible by $9$.
