Let $ A$, $ B$ be the number of digits of $ 2^{1998}$ and $ 5^{1998}$ in decimal system. $ A \plus{} B \equal{} ?$

$\textbf{(A)}\ 1998  \qquad\textbf{(B)}\ 1999  \qquad\textbf{(C)}\ 2000  \qquad\textbf{(D)}\ 3996  \qquad\textbf{(E)}\ 3998$
1. To find the number of digits of \(2^{1998}\) and \(5^{1998}\), we use the formula for the number of digits of a number \(n\), which is given by:
   \[
   \text{Number of digits of } n = \left\lfloor \log_{10} n \right\rfloor + 1
   \]
   Therefore, the number of digits of \(2^{1998}\) is:
   \[
   A = \left\lfloor \log_{10} (2^{1998}) \right\rfloor + 1
   \]
   Similarly, the number of digits of \(5^{1998}\) is:
   \[
   B = \left\lfloor \log_{10} (5^{1998}) \right\rfloor + 1
   \]

2. Using the properties of logarithms, we can simplify the expressions for \(A\) and \(B\):
   \[
   \log_{10} (2^{1998}) = 1998 \log_{10} 2
   \]
   \[
   \log_{10} (5^{1998}) = 1998 \log_{10} 5
   \]

3. Adding these two logarithmic expressions, we get:
   \[
   \log_{10} (2^{1998}) + \log_{10} (5^{1998}) = 1998 \log_{10} 2 + 1998 \log_{10} 5
   \]
   Using the property of logarithms that \(\log_{10} (a \cdot b) = \log_{10} a + \log_{10} b\), we have:
   \[
   1998 \log_{10} 2 + 1998 \log_{10} 5 = 1998 (\log_{10} 2 + \log_{10} 5) = 1998 \log_{10} (2 \cdot 5) = 1998 \log_{10} 10 = 1998
   \]

4. Since \(\log_{10} (2^{1998}) + \log_{10} (5^{1998}) = 1998\), we can write:
   \[
   \left\lfloor \log_{10} (2^{1998}) \right\rfloor + \left\lfloor \log_{10} (5^{1998}) \right\rfloor \leq 1998 - 1
   \]
   This is because the sum of the floors of two numbers is less than or equal to the floor of their sum plus one.

5. Therefore, the number of digits \(A\) and \(B\) can be expressed as:
   \[
   A = \left\lfloor \log_{10} (2^{1998}) \right\rfloor + 1
   \]
   \[
   B = \left\lfloor \log_{10} (5^{1998}) \right\rfloor + 1
   \]

6. Adding \(A\) and \(B\), we get:
   \[
   A + B = \left\lfloor \log_{10} (2^{1998}) \right\rfloor + 1 + \left\lfloor \log_{10} (5^{1998}) \right\rfloor + 1
   \]
   \[
   A + B = \left\lfloor \log_{10} (2^{1998}) \right\rfloor + \left\lfloor \log_{10} (5^{1998}) \right\rfloor + 2
   \]

7. Since \(\left\lfloor \log_{10} (2^{1998}) \right\rfloor + \left\lfloor \log_{10} (5^{1998}) \right\rfloor = 1998 - 1\), we have:
   \[
   A + B = 1998 - 1 + 2 = 1999
   \]

The final answer is \(\boxed{1999}\).