public class Exercise26 { public static void main(String arg[]) { // Checksum, Q1.3.33 // The International Standard Book Number (ISBN) is a 10-digit code that uniquely // specifies a book. The rightmost digit is a checksum digit that can be uniquely // determined from the other 9 digits, from the condition that // d1 + 2d2 +3d3 +... + 10d10 must be a multiple of 11 (here di denotes the // ith digit from the right). The checksum digit di can be any value from 0 to 10. // The ISBN convention is to use the character 'X' to denote 10. // Example: the checksum digit corresponding to 020131452 is 5 since 5 is the only // value of s between 0 and 10 for which // 10•0 + 9•2 + 8•0 + 7•1 + 6•3 + 5•1 + 4•4 + 3•5 + 2•2 + 1•s // is a multiple of 11. // Write a program that takes a 9-digit integer from user input, // computes the checksum digit, and prints out the complete ISBN number. // Note that the right-most digit of the user input is actually d2 in the // formula above. System.out.print("Enter a partial ISBN (first 9 digits): "); String partialISBN = StdIn.readString(); int N = Integer.parseInt(partialISBN)); int sum = 0; // TODO: Add code here to compute the sum for the first 9 digits: // 2*digit1 +3*digit2 +... + 10*digit9 // Hint: Use digit = N % 10 to get the right-most digit of N. // of N. And you can use N = N / 10 to discard the last digit of N // (and also shifting all the digits over one place to the right). // Next, figure out how much extra needs to be added to the sum // so that it would be a multiple of 11. This is the value of the // checksum digit. int extra; if ((sum % 11) == 0) extra = 0; // no extra needed, because sum is a multiple of 11 already else extra = 11 - (sum % 11); // a little extra needed // TODO: add code here to decide what the last digit should be. If the // extra amount is 10, then the last digit is "X". Otherwise, the last // digit is just whatever the extra amount is. String last; // Finally, print the full ISBN with checksum. System.out.println("The full ISBN number is " + partialISBN + last); } }