we claim that $n=13$ is the answer. first note that if $p|x$ , then $p|z$ and $p|y$. now we want $n$ such that: $$v_p(x)+v_p(y)+v_p(z) \le n v_p(x+y+z)$$$WLOG$ $ v_p(x)$ is minimum . we know that $v_p(z) \le 3v_p(x)$ and $v_p(y)\le 9 v_p(x)$. therefore $ v_p(x)+v_p(y)+v_p(z) \le 13 v_p(x)=13v_p(x+y+z)$ now put ordered$(x,y,z)=(p,p^9,p^3)$ to see that $n\ge 13$ so its done.