<s>[INST] How do I prove that $f(n) = 0$ for all $n$ where $n$ is a positive integer? In particular, $f$ is defined such that $f(mn)=f(m)+f(n)$, where $m$ and $n$ are positive integers. Also $f(n)\ge 0$ for all $n$, $f(10)=0$, and $f(n)=0$ if $n$ ends in a $3$. Thank you. 
 [/INST] *

*If $n\equiv 1\pmod{10}$ then $3n\equiv 3\pmod{10}$ and so $f(n)=f(3n)-f(3)=0$.

*If $n\equiv 7\pmod{10}$ then $3n\equiv 1\pmod{10}$ and so $f(n)=f(3n)-f(3)=0$.

*If $n\equiv 9\pmod{10}$ then $3n\equiv 7\pmod{10}$ and so $f(n)=f(3n)-f(3)=0$.


From $f(2)\ge 0$ and $f(5)\ge 0$ and $f(2)+f(5)=f(10)=0$, we obtain $f(2)=f(5)=0$.
Now let $n\in\Bbb N$ be arbitrary. Write $n=2^r5^su$ with $\gcd(u,10)=1$. Then $u\equiv 1,3,7,9\pmod{10}$ and hence $f(n)=rf(2)+sf(5)+f(u)=0$.
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