For each positive integer $n$, an non-negative integer $f(n)$ is associated in such a way that the following three rules are satisfied:
i) $f(a b)=f(a)+f(b)$.

ii) $f(n)=0$ if $n$ is a prime greater than 10.

iii) $f(1)<f(243)<f(2)<11$.

Knowing that $f(2106)<11$, determine the value of $f(96)$.
Solution

By property $i$), we have $f(243)=f\left(3^{5}\right)=5 f(3)$. Given that

$$
0 \leq f(1)<5 f(3)<f(2)<11
$$

and that $5 f(3)$ is a multiple of 5, we have $5 f(3)=5$, that is, $f(3)=1$. Note that $2106=2 \cdot 3^{4} \cdot 13$. Thus, by property $i$),

$$
f(2106)=f(2)+4 f(3)+f(13)=f(2)+4
$$

From $f(2106)<11$, it follows that $f(2)<7$. Using $i$ i), we have $5<f(2)<7$ and thus $f(2)=6$. Therefore,

$$
f(96)=f\left(3 \cdot 2^{5}\right)=f(3)+5 f(2)=31
$$