Books are identified by an International Standard Book Number (ISBN), a 10-digit code x1 x2…x10, assigned by the publisher. These 10 digits consist of blocks identifying the language, the publisher, the number assigned to the book by its publishing company, and finally, a 1-digit check digit that is either a digit or the letter X (used to represent 10). This check digit is selected so that \sum_{i=1}^{10} ix_i \equiv 0 \text{ mod } 11 and is used to detect errors in individual digits and transposition of digits.
a. The ISBN for a book is 1-32-794182- Q} where Q is the check digit. What is Q?
b. The ISBN of another book is 2-429-39M18-6. Find the digit M.
Answer:
[tex]Q = 1[/tex]
[tex]M = 2[/tex]
Step-by-step explanation:
Solving (a):
Given: 1-32-794182- Q
Required: Find Q
To do this, we first multiply each digit in descending order of 10.
So, we have:
[tex]Product = 1 * 10 + 3 * 9 + 2 * 8 + 7 * 7 + 9 * 6 + 4 * 5 + 1 * 4 + 8 * 3 + 2 * 2 + Q * 1[/tex]
[tex]Product = 208 + Q *1[/tex]
[tex]Product = 208 + Q[/tex]
Equate the modulus of the product by 11 to 0
[tex]Product \% 11 = 0[/tex]
[tex](208 + Q) \% 11 = 0[/tex]
The next number greater than 208 divisible by 11 is 209.
So, we have:
[tex](208 + Q) \% 11 = 209 \% 11[/tex]
Express 209 as 208 + 1
[tex](208 + Q) \% 11 = (208 + 1)\% 11[/tex]
By comparison:
[tex]Q = 1[/tex]
Hence, the checksum is 1
Solving (a):
Given: 2-429-39M18-6
Required: Find M
Apply the same step in (a)
So, we have:
[tex]Product = 2 * 10 + 4 * 9 + 2 * 8 + 9 * 7 + 3 * 6 + 9 * 5 + M * 4 + 1 * 3 + 8 * 2 + 6 * 1[/tex]
[tex]Product = 198 + 4M + 25[/tex]
[tex]Product = 223+ 4M[/tex]
Equate the modulus of the product by 11 to 0
[tex]Product \% 11 = 0[/tex]
[tex](223+ 4M) \% 11 = 0[/tex]
The next number greater than 223 divisible by 11 is 231.
So, we have:
[tex](223+ 4M) \% 11 = 231 \% 11[/tex]
Express 231 as 223 + 8
[tex](223+ 4M) \% 11 = (223 + 8) \% 11[/tex]
By comparison:
[tex]4M = 8[/tex]
Divide both sides by 4
[tex]M = 2[/tex]