The number $2^{1997}$ has $m$ decimal digits, while the number $5^{1997}$ has $n$ digits. Evaluate $m+n$.
1. To find the number of decimal digits of a number \( a^b \), we use the formula:
   \[
   \text{Number of digits} = \lfloor 1 + b \log_{10}(a) \rfloor
   \]
   where \( \lfloor x \rfloor \) denotes the floor function, which gives the greatest integer less than or equal to \( x \).

2. First, we calculate the number of digits \( m \) in \( 2^{1997} \):
   \[
   m = \lfloor 1 + 1997 \log_{10}(2) \rfloor
   \]
   We need the value of \( \log_{10}(2) \). Using a calculator or logarithm table, we find:
   \[
   \log_{10}(2) \approx 0.3010
   \]
   Substituting this value in, we get:
   \[
   m = \lfloor 1 + 1997 \cdot 0.3010 \rfloor
   \]
   \[
   m = \lfloor 1 + 600.097 \rfloor
   \]
   \[
   m = \lfloor 601.097 \rfloor
   \]
   \[
   m = 601
   \]

3. Next, we calculate the number of digits \( n \) in \( 5^{1997} \):
   \[
   n = \lfloor 1 + 1997 \log_{10}(5) \rfloor
   \]
   We need the value of \( \log_{10}(5) \). Using a calculator or logarithm table, we find:
   \[
   \log_{10}(5) \approx 0.6990
   \]
   Substituting this value in, we get:
   \[
   n = \lfloor 1 + 1997 \cdot 0.6990 \rfloor
   \]
   \[
   n = \lfloor 1 + 1396.803 \rfloor
   \]
   \[
   n = \lfloor 1397.803 \rfloor
   \]
   \[
   n = 1397
   \]

4. Finally, we add the number of digits \( m \) and \( n \):
   \[
   m + n = 601 + 1397 = 1998
   \]

The final answer is \(\boxed{1998}\)