smc

For positive integers $n$, ${\log }_{8}n$ is rational if and only if $n=2^k$ for an integer $k$. That's because ${\log }_8{2}^k=k{\log }_{8}2=\frac{k}{3}$. So we actually want to find ${\sum }_{k=0}^{10}{\log }_8{2}^k$, since $2^{11}$ will be over $1997$. Using log properties, we get ${\sum }_{k=0}^{10}k{\log }_{8}2$ $\frac{1}{3}{\sum }_{k=0}^{10}k$ $\frac{1}{3}\cdot (\frac{10\cdot 11}{2})$ $\frac{55}{3}$, and the answer is $\boxed{\frac{55}{3}}$