Question 26
Let the function $f: N \rightarrow R$ be such that for every natural number $n \gt 1$, there is a prime divisor $p$ of $n$ such that:
$$
f(n) = f\left(\frac{n}{p}\right) - f(p)
$$
If $f(2^{2007}) + f(3^{2008}) + f(5^{2009}) = 2006$, compute $f(2007^2) + f(2008^3) + f(2009^5)$
(adapted).
$$
f(n) = f\left(\frac{n}{p}\right) - f(p)
$$
If $f(2^{2007}) + f(3^{2008}) + f(5^{2009}) = 2006$, compute $f(2007^2) + f(2008^3) + f(2009^5)$
(adapted).