Editing ISBN (section)

===ISBN-10 check digit calculation===
<!-- "Damm algorithm" links here -->
Each of the first nine digits of the 10-digit ISBN—excluding the check digit itself—is multiplied by its (integer) weight, descending from 10 to 2, and the sum of these nine products found. The value of the check digit is simply the one number between 0 and 10 which, when added to this sum, means the total is a multiple of 11.

For example, the check digit for an ISBN-10 of 0-306-40615-''?'' is calculated as follows:
{{block indent|<math>
\begin{align}
 s &= (0\times 10)+(3\times 9)+(0\times 8)+(6\times 7)+(4\times 6)+(0\times 5)+(6\times 4)+(1\times 3)+(5\times 2)\\
   &= 130.
\end{align}
</math>}}
Adding 2 to 130 gives a multiple of 11 (because 132 = 12×11)—this is the only number between 0 and 10 which does so. Therefore, the check digit has to be 2, and the complete sequence is <nowiki>ISBN 0-306-40615-2</nowiki>. If the value of <math>x_{10}</math> required to satisfy this condition is 10, then an 'X' should be used.

Alternatively, [[modular arithmetic]] is convenient for calculating the check digit using modulus 11. The [[remainder]] of this sum when it is divided by 11 (i.e. its value modulo 11), is computed. This remainder plus the check digit must equal either 0 or 11. Therefore, the check digit is (11 minus the remainder of the sum of the products modulo 11) modulo 11. Taking the remainder modulo 11 a second time accounts for the possibility that the first remainder is 0. Without the second modulo operation, the calculation could result in a check digit value of {{nobr|11 − 0 {{=}} 11}}, which is invalid. (Strictly speaking, the ''first'' "modulo 11" is not needed, but it may be considered to simplify the calculation.)

For example, the check digit for the <nowiki>ISBN</nowiki> of 0-306-40615-''?'' is calculated as follows:
{{block indent|<math>
\begin{align}
 s &= (11 - ( ( (0\times 10)+(3\times 9)+(0\times 8)+(6\times 7)+(4\times 6)+(0\times 5)+(6\times 4)+(1\times 3)+(5\times 2) ) \,\bmod\, 11 ) ) \,\bmod\, 11\\
   &=    (11 - ( (0 + 27 +   0 +  42 +  24 +   0 + 24  +   3 + 10 ) \,\bmod\, 11) ) \,\bmod\, 11\\
   &= (11-((130) \,\bmod\, 11))\,\bmod\, 11 \\
   &= (11-(9))\,\bmod\, 11 \\
   &= 2\,\bmod\, 11 \\
   &= 2
\end{align}
</math>}}

Thus the check digit is 2.

It is possible to avoid the multiplications in a software implementation by using two accumulators. Repeatedly adding <code>t</code> into <code>s</code> computes the necessary multiples:
<syntaxhighlight lang="C">
// Returns ISBN error syndrome, zero for a valid ISBN, non-zero for an invalid one.
// digits[i] must be between 0 and 10.
int CheckISBN(int const digits[10]) {
  int i, s = 0, t = 0;

  for (i = 0; i < 10; ++i) {
    t += digits[i];
    s += t;
  }
  return s % 11;
}
</syntaxhighlight>
The modular reduction can be done once at the end, as shown above (in which case <code>s</code> could hold a value as large as 496, for the invalid <nowiki>ISBN 99999-999-9-X</nowiki>), or <code>s</code> and <code>t</code> could be reduced by a conditional subtract after each addition.