X OBM

$f(30)=f(2)+f(3)+f(5)=0$ and $f(n)$ is non-negative, so $f(2)=f(3)=f(5)=0$. For any positive integer $n$ not divisible by $2$ or $5$ we can find a positive integer $m$ such that $mn\equiv 7(\mathrm{mod}10)$. But then $f(mn)=0$, so $f(n)=0$. It is a trivial induction that $f(2^a{5}^{b}n)=f(5^{b}n)=f(n)$, so $f$ is identically zero.