G4.3 Let $n$ and $\frac{47}{5}\left(\frac{4}{47}+\frac{n}{141}\right)$ be positive integers. If $r$ is the remainder of $n$ divided by 15 , find the value of $r$.
$\frac{47}{5}\left(\frac{4}{47}+\frac{n}{141}\right)=\frac{4}{5}+\frac{n}{15}=\frac{n+12}{15}$, which is an integer $n+12=15 k$, where $k$ is a positive integer $r=3$