1's and 0's in Binary system

$55 _ { ( 10 ) } = 110111 _ { ( 2 ) }$

$99 _ { ( 10 ) } = 1100011 _ { ( 2 ) }$

$77 _ { ( 10 ) } = 1001101 _ { ( 2 ) }$

$66 _ { ( 10 ) } = 1000010 _ { ( 2 ) }$

$37 _ { ( 10 ) } = 100101 _ { ( 2 ) }$

$55$ has $5$ 1's, and $1$ 0's, so $55$ is not the answer.

$99$ has $4$ 1's, and $3$ 0's, so $99$ is also not the answer.

$77$ has $4$ 1's, and $3$ 0's, so $77$ is not the answer, either.

$66$ has $2$ 1's, and $5$ 0's, so $66$ is the answer.

$37$ has $3$ 1's, and $3$ 0's, so $37$ is not the answer, too.

$66 = 2^{6} + 2 = 1000000_{2} + 10_{2} = 1000010_{2}$

The way to go about it is to look for the number which is just a little larger than a power of 2.

1

bin(66)

Gives:

1

0b1000010