Do it in quaternary. Add pairs of bits.

Or you can form the alternating sum of the bits, e.g. for 0b10011001 you calculate 1-0+0-1+1-0+0-1 = 0 which is divisible by three. (That's similar to the divisibility test by 11 of a number in base-10, or more generally testing if a number in base `b` is divisible by b+1)