= Solution
Let $M(x)=ax^6+bx^5+cx^4+dx^3+ex^2+fx+g$. 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> are
$$
\boxed{\langle a,b,c,d,e,f,g,\ a\mathbin\oplus b\mathbin\oplus c\mathbin\oplus d\mathbin\oplus e\mathbin\oplus f\mathbin\oplus g\rangle.}
$$
They 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. Therefore
$$
\boxed{\text{every odd number of bit flips is detected; every even number is undetected}.}
$$
In 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>.
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