3. The function $f(x)$ is such that for all natural $n>1$ there exists a prime divisor $p$ of the number $n$ such that

$$
f(n)=f\left(\frac{n}{p}\right)-f(p)
$$

It is known that $f(1001)=1$. What is $f(1002) ?$
Solution. Note that for any prime number p, the value $f(p)=f(1)-$ $f(p)$. Therefore, $f(p)=\frac{f(1)}{2}$ for any prime number. For prime numbers p and q, we get that either $\mathrm{f}(\mathrm{pq})=\mathrm{f}(\mathrm{p})-\mathrm{f}(\mathrm{q})=0$, or $\mathrm{f}(\mathrm{pq})=\mathrm{f}(\mathrm{q})-\mathrm{f}(\mathrm{p})=0$. For three prime numbers $\mathrm{p}, \mathrm{q}$, and $\mathrm{r}$, we get that $\mathrm{f}(\mathrm{pqr})=\mathrm{f}(\mathrm{pq})-\mathrm{f}(\mathrm{r})=-\mathrm{f}(\mathrm{r})=-\frac{\mathrm{f}(1)}{2}$ (the order of the prime numbers can be different). Then $f(1001)=f(7 \cdot 11 \cdot 13)=-\frac{f(1)}{2}=1$. But then $f(1002)=f(2 \cdot 3 \cdot 167)=-\frac{f(1)}{2}=1$.

Answer: 1.

Criteria: Full solution - 7 points. It is shown that $f(1)=-2$ or there is a proof that $f(p q)=0 - 2$ points.