For a valid 10-digit ISBN number, $x_{10}−x_9x_8x_7−x_6x_5x_4x_3x_2−x_1$, if we switch 2 digit, $x_{10}−x_9x_8x_2−x_6x_5x_4x_3x_7−x_1$, then it becomes a invalid number. But for a valid 13-digit ISBN number, if I switch adjacent digit, then the number becomes invalid.
For example:
$978-0-8218-5209-5$, Advanced Calculus 2nd edition by Patrick M. Fitzpatrick
Switch $x_2=9$ and $x_6=8$, we have $978-0-8219-5208-5$, then apply the check formula, we have \begin{align*} \sum_{i=1}^{6}x_{2i+1}&=0+5+1+8+8+9=31\\ 3\sum_{i=1}^{6}x_{2i}&=3(8+2+9+2+0+7)=84 \end{align*} So, $-32-84=-115\equiv-15\equiv5\pmod{10}$, we still have $x_1=5$.
Switch $x_2=9$ to $x_3=0$ we have $978-0-8219-5280-5$, then apply the check formula, we have \begin{align*} \sum_{i=1}^{6}x_{2i+1}&=8+5+1+8+8+9=39\\ 3\sum_{i=1}^{6}x_{2i}&=3(0+2+9+2+0+7)=60 \end{align*} Thus, $-39-60=-99\equiv-9\equiv1\pmod{10}$; hence, we receive $x_1=1$ which is a different digit and isn't a valid code.
Can someone tell me why ISBN-13 does not always detect a transposition error as ISBN-10, just detect adjacent transposition error? Thanks.
If the first twelve digits are $x_1, x_2, \dotsc, x_{12}$, then the check digit $x_{13}$ is chosen so that \begin{equation*} x_1 + 3x_2 + x_3 + 3x_4 + \cdots + x_{11} + 3x_{12} + x_{13}\equiv 0\mod{10}. \end{equation*} Clearly if any two even-position digits, or any two odd-position digits, are transposed, then the code will still look correct.
Note that the ISBN-13 check digit system will not even detect adjacent transposition errors if the difference between the adjacent digits is divisible by $5$, since \begin{equation*} (x_i+3x_{i+1})-(x_{i+1}+3x_i) = 2(x_{i+1}-x_i), \end{equation*} which is $\equiv 0\mod{10}$ if $x_{i+1}-x_i\equiv 0\mod{5}$.