 Solution

Let $M(x)=a{x}^6+b{x}^5+c{x}^4+d{x}^3+e{x}^2+fx+g$. With CRC generator polynomial $x+1$, the appended check has one bit. The preceding polynomial remainder rule gives $(xM)(1)=M(1)$, so put $h=a+b+c+d+e+f+g$ in the field with two elements. The eight transmitted bits areThey have even Hamming weight, so this cyclic redundancy check is a single parity-check code. For a substitution error polynomial $E$, detection occurs exactly when $E(1)=1$: an odd number of the eight bits, including the check bit, have flipped. ThereforeIn particular, every single-bit error is detected, while flipping any two distinct bits passes this check. This conclusion assumes that the eight-bit word's boundaries and length are retained; it supplies no general protection against insertions, deletions or lost framing.