Let us prove a stronger statement: $5 \not | n \implies f(n) = 0$. $f(6) = f(2) + f(3) = 0$ $f(4) = 2 f(2) = 0$ $f(2) = f(3) = 0$ Lemma: Given an odd integer $k, 5 \not | k$, then some $n$ exists such that $2^n k \equiv 4 \bmod 10$. Proof: Clearly $2^n k$ is even, so we only require $\equiv 2 \bmod 5$. If $k \not \equiv 0 \bmod 5$ then such $n$ obviously exists. From our lemma, for such $n, k$ we have $f(2^n k) = n f(2) + f(k) = f(k) = 0$ From which our claim follows. Therefore, $f(1998) = \boxed{0}$.