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 $\boxed{1999}$.