Non-negative $f(n)$ satisfies $f(ab)=f(a)+f(b)$, $f(n)=0$ for $n$ a prime greater than $10$, and $f(1)
For each positive integer n, a non-negative integer $f (n)$ is associated so that the following three rules are met:
i) $f (ab) = f (a)+ f (b).$
ii) $f (n) = 0 $ if $n$ is a greater prime than $10$.
iii) $f (1) < f (243) < f (2) < 11$
Knowing that $f(2106) < 11$, determine the value of $f(96)$.
By property i), we have $f (243) = f (3^5) = 5f (3).$ Given that
$0 ≤ f (1) < 5f (3) < f (2) < 11$, and that $5f (3)$ is a multiple of $5$, we have $5f (3) = 5$, that is, $f (3) = 1$. Note that $2106 = 2·3^4·13.$
So, by property i),
$f(2106)=f(2)+4f(3)+f(13)=f(2)+4.$

You now know that $f(2)+4<11$ and $f(2)>5$; it then follows from $f$'s integrality that $f(2)=6$. Then $f(96)=5f(2)+f(3)=5\cdot6+1=31$.
