$33 \cdot 72$ If $D(n)$ represents the number of pairs of adjacent different digits in the binary representation of a positive integer $n$. For example, $D(3)=D\left(11_{2}\right)=0, D(21)=D\left(10101_{2}\right)=4$, $D(97)=D\left(1100001_{2}\right)=2$, etc. The number of positive integers $n$ not greater than 97 that satisfy $D(n)=2$ is
(A) 16.
(B) 20.
(C) 26.
(D) 30.
(E) 35.
(48th American High School Mathematics Examination, 1997)
[Solution] Considering $97<2^{7}=128$, it is evident that the binary representation of the above $n$ does not exceed 7 digits.

Also, $97=2^{6}+2^{5}+1=1100001_{2}$. Thus, for $n$ not exceeding 97 and satisfying $D(n)=2$, their binary representations are as follows:
7 digits: $1100001_{2}, 1011111_{2}, 1001111_{2}, 1000111_{2}, 1000011_{2}$,
$1000001_{2}$;
6 digits: $111101_{2}, 111011_{2}, 111001_{2}, 110111_{2}, 110011_{2}$,
$110001_{2}, 101111_{2}, 100111_{2}, 100011_{2}, 100001_{2}$;
5 digits: $11101_{2}, 11011_{2}, 11001_{2}, 10111_{2}, 10011_{2}, 10001_{2}$;
4 digits: $1101_{2}, 1011_{2}, 1001_{2}$;
3 digits: $101_{2}$.
In total, there are 26.
Therefore, the answer is $(C)$.