Let us look on such an equality:
$$f(1000) = f(10^3) = 3 \cdot f(10)$$
Let us show now that for any integer $k$ $f(1000)$ can equal $3k$. Let $\operatorname{ord}(a)$ be a maximum degree of $2$, by what $a$ is divisible. Let
$$a = 2^i(2a_0 + 1), \quad b = 2^m(2b_0 + 1). \text{ Thus}$$
$$\operatorname{ord}(ab) = \operatorname{ord}(2^{i+m}(2a_0+1)(2b_0+1)) = i+m = \operatorname{ord}(a)+\operatorname{ord}(b). \text{ Let } f(a)=k \cdot \operatorname{ord}(a), \text{ then}$$
$$f(ab) = k \cdot \operatorname{ord}(ab) = k \cdot (\operatorname{ord}(a) + \operatorname{ord}(b)) = k \cdot \operatorname{ord}(a) + k \cdot \operatorname{ord}(b) = f(a) + f(b)$$
$$\Rightarrow f(1000) = k \cdot \operatorname{ord}(2^3 \cdot 125) = 3k,$$
That is what we had to prove.