Define $ \{ a_n \}_{n\equal{}1}$ as follows: $ a_1 \equal{} 1989^{1989}; \ a_n, n > 1,$ is the sum of the digits of $ a_{n\minus{}1}$. What is the value of $ a_5$?
1. **Establishing the congruence relationship:**
   - Given \( a_1 = 1989^{1989} \), we need to find \( a_5 \).
   - Note that \( a_n \) is the sum of the digits of \( a_{n-1} \) for \( n > 1 \).
   - By properties of digit sums, \( a_{n-1} \equiv a_n \pmod{9} \). Therefore, \( a_1 \equiv a_2 \equiv a_3 \equiv a_4 \equiv a_5 \pmod{9} \).

2. **Calculating \( a_1 \mod 9 \):**
   - \( 1989 \equiv 1 + 9 + 8 + 9 = 27 \equiv 0 \pmod{9} \).
   - Thus, \( 1989^{1989} \equiv 0^{1989} \equiv 0 \pmod{9} \).
   - Therefore, \( a_1 \equiv 0 \pmod{9} \).

3. **Bounding the values of \( a_n \):**
   - \( 1989^{1989} < 10000^{1989} \).
   - \( \log_{10}(10000^{1989}) = 4 \times 1989 = 7956 \).
   - Hence, \( 10000^{1989} \) has 7957 digits, so \( 1989^{1989} \) has at most 7957 digits.

4. **Calculating the maximum possible values for \( a_2, a_3, a_4 \):**
   - If \( a_1 \) has 7957 digits, the maximum sum of its digits (i.e., \( a_2 \)) is \( 9 \times 7957 = 71613 \).
   - The maximum sum of the digits of \( a_2 \) (i.e., \( a_3 \)) is when \( a_2 = 69999 \), so \( a_3 \leq 6 + 9 + 9 + 9 + 9 = 42 \).
   - The maximum sum of the digits of \( a_3 \) (i.e., \( a_4 \)) is when \( a_3 = 39 \), so \( a_4 \leq 3 + 9 = 12 \).

5. **Determining \( a_5 \):**
   - The maximum sum of the digits of \( a_4 \) (i.e., \( a_5 \)) is when \( a_4 = 12 \), so \( a_5 \leq 1 + 2 = 3 \).
   - Since \( a_5 \equiv 0 \pmod{9} \) and \( a_5 \) must be a single-digit number, the only possible value is \( 9 \).

Conclusion:
\[
a_5 = \boxed{9}
\]