6. If a binary number $n=\left(a_{m} a_{m-1} \cdots a_{1} a_{0}\right)_{2}$ satisfies $\left(a_{m} a_{m-1} \cdots a_{1} a_{0}\right)_{2}=$ ( $\left(a_{0} a_{1} \cdots a_{m-1} a_{m}\right)_{2}$, then $n$ is called a "binary palindrome". How many "binary palindrome" numbers are there among the natural numbers not exceeding 1988?
6. The number of $K$-bit binary palindromes is $2^{\left[\frac{1}{2}(K-1)\right]}$, and $1988=(11111000100)_{2}$ is an 11-bit binary number. There are only two 11-bit binary palindromes greater than 1988, which are $(11111011111)_{2}$ and $(11111111111)_{2}$. Therefore, the number of binary palindromes sought is
$$
\sum_{K=1}^{11} 2^{\left[\frac{1}{2}(K-1)\right]}-2=2\left(1+2+2^{2}+2^{3}+2^{4}\right)+2^{5}-2=92 \text { (numbers). }
$$