The function $\mathrm{f}(\mathrm{n})$ is defined on the positive integers and takes non-negative integer values. It satisfies (1) $f(m n)=f(m)+f(n),(2) f(n)=0$ if the last digit of $n$ is 3, (3) $f(10)=0$. Find $\mathrm{f}(1985)$.
If $\mathrm{f}(\mathrm{mn})=0$, then $\mathrm{f}(\mathrm{m})+\mathrm{f}(\mathrm{n})=0$ (by $(1))$. But $\mathrm{f}(\mathrm{m})$ and $\mathrm{f}(\mathrm{n})$ are non-negative, so $\mathrm{f}(\mathrm{m})=\mathrm{f}(\mathrm{n})=0$. Thus $f(10)=0$ implies $f(5)=0$. Similarly $f(3573)=0$ by (2), so $f(397)=0$. Hence $f(1985)$ $=\mathrm{f}(5)+\mathrm{f}(397)=0$.