To each positive integer $ n$ it is assigned a non-negative integer $f(n)$ such that the following conditions are satisfied:

(1) $ f(rs) \equal{} f(r)\plus{}f(s)$
(2) $ f(n) \equal{} 0$, if the first digit (from right to left) of $ n$ is 3.
(3) $ f(10) \equal{} 0$.

Find $f(1985)$. Justify your answer.
1. We start by analyzing the given conditions:
   - \( f(rs) = f(r) + f(s) \)
   - \( f(n) = 0 \) if the first digit (from right to left) of \( n \) is 3.
   - \( f(10) = 0 \)

2. From condition (3), we know \( f(10) = 0 \).

3. Consider \( f(10k + r) \). Since \( 10k + r \) is a number where \( r \) is the last digit, we can write:
   \[
   f(10k + r) = f(10) + f(k) + f(r) = 0 + f(k) + f(r) = f(k) + f(r)
   \]
   However, if \( r \) is 3, then \( f(10k + 3) = 0 \) by condition (2). This implies \( f(k) + f(3) = 0 \). Since \( f(3) = 0 \), we have \( f(k) = 0 \).

4. We need to show that \( f(n) = 0 \) for all \( n \). We start by proving it for the digits 1 through 9:
   - \( f(1) \): From \( f(10) = f(1) + f(10) = 0 \), we get \( f(1) = 0 \).
   - \( f(2) \): From \( f(10) = f(2) + f(5) = 0 \), we get \( f(2) = 0 \) and \( f(5) = 0 \).
   - \( f(4) \): From \( f(4) = f(2) + f(2) = 0 + 0 = 0 \).
   - \( f(6) \): From \( f(6) = f(2) + f(3) = 0 + 0 = 0 \).
   - \( f(8) \): From \( f(8) = f(2) + f(2) + f(2) + f(2) = 0 + 0 + 0 + 0 = 0 \).
   - \( f(7) \): From \( f(35) = f(5) + f(7) = 0 \), we get \( f(7) = 0 \).
   - \( f(9) \): From \( f(9) = f(3) + f(3) = 0 + 0 = 0 \).

5. Since \( f(n) = 0 \) for all single-digit numbers, we can extend this to all positive integers by induction. Assume \( f(k) = 0 \) for all \( k \) less than \( n \). For \( n \), write \( n = 10a + b \) where \( b \) is the last digit. Then:
   \[
   f(n) = f(10a + b) = f(10a) + f(b) = 0 + 0 = 0
   \]
   Thus, \( f(n) = 0 \) for all \( n \).

6. Specifically, for \( f(1985) \):
   \[
   f(1985) = 0
   \]

The final answer is \( \boxed{0} \).