Need help with error detecting code calculation!! A University's student ID numbers are examples of error detecting codes. Each
student ID has 9 digits, with the ninth digit being the check digit. 
The check vector is
$c = (9, 8, 7, 6, 5, 4, 3, 2, 1)$,
and the check digit is such that if $x$ is a valid student ID then
$c · x = 0$ in Z11.


*

*Show that there is an error in the student ID $430351602$.

*A common error when entering an SID on machine-readable forms is to
substitute $n + 1$ for n, so a $0$ gets entered as a $1$, or $1$ as $2$, for example.
Assuming that the error in the SID $430351602$ is exactly one error of this
type, ﬁnd the correct SID.


Please help with how to calculate this!

To see if $\langle\,x_1,x_2, \dots,x_9\,\rangle$ is a valid ID, you multiply the terms with the corresponding terms in the check vector, like this
$$\begin{array}{lccccccccc}
\mathbf{c}:\ &  9 & 8  & 7 & 6  & 5  & 4 & 3  & 2 & 1\\
\mathbf{x}:\ &  4 & 3  & 0 & 3  & 5  & 1 & 6  & 0 & 2\\
             & 36 & 24 & 0 & 18 & 25 & 4 & 18 & 0 & 2
\end{array}$$
then add the products: $36+24+0+18+25+4+18+0+2=127$. This number yields a remainder of 6 when divided by 11; in other words, it's not $0$ in $\mathbb{Z}_{11}$, so the ID as entered must be incorrect.
Assuming that the entry error came from entering some $x_i$ as one more than it should have been, it must be that the intended correct ID must have had exactly one entry that was supposed to be $x_i-1$ but was entered as $x_i$. Thus, the intended ID must have been such that
$$
c_1x_1+c_2x_2+\cdots+c_i(x_i-1)+\cdots+c_9x_9
$$
had a remainder of zero when divided by 11. We saw above that the sum 
$$
c_1x_1+c_2x_2+\cdots+c_ix_i+\cdots+c_9x_9
$$
had a remainder of 6 when divided by 11, so the error must have had one $c_ix_i$ with remainder of 6 when divided by 11, which is to say that $c_ix_i=11k+6$ for some $k$. But then in the correct ID we'd have 
$$
c_i(x_i-1)=c_ix_i-c_i=(11k+6)-c_i=11k+(6-c_1)
$$
and this could only be a multiple of 11 if $6-c_1=0$, which is to say $c_i=6$. In other words, looking at the table above, the error must have been in the column where $c_i=6$ so the error must have been entering 3 in that column, rather than the intended 2, so the correct ID should be
$$
4\ 3\ 0\ \color{red}2\ 5\ 1\ 6\ 0\ 2
$$
