Hexadecimal format is given
0 x 27 = ( 0010 0111 )2
0 x 3f = ( 0011 1111 )2
ab & 3f = 27
ab & ( 0011 1111 ) = ( 0010 0111 )
a & ( 0011 ) = ( 0010 )
b & ( 1111 ) = ( 0111 )
if a = ( 0010 ), then a & ( 0011 ) will give (0010 )
if b = ( 0111 ), then b & ( 1111 ) will give ( 0111 )
so the value of a and b are ( 0010) and (0111)
ab = (0010 0111) = (0 x 27 ) 16