19. How many 2-digit numbers can be written as the sum of exactly six different powers of 2, including $2^{0}$ ?
A 0
B 1
С 2
D 3
E 4
19. C Since $2^{7}=128$ is larger than 100 , the only powers we can choose from are the first seven powers: $2^{0}$ to $2^{6}$, i.e. $1,2,4,8,16,32,64$. The sum of all seven is 127 . We need to eliminate one of these, and be left with a total under 100 , so the only possibilities to remove would be 32 or 64 . Hence there are two options:
$$
1+2+4+8+16+32=63 \text {, and } 1+2+4+8+16+64=95 \text {. }
$$