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Law Incorporation [45]
1 year ago
13

The following checksum formula is widely used by banks and credit card companies to validate legal account numbers: d0 + f(d1) +

d2 + f(d3) + d4 + f(d5) + . . . = 0 (mod 10) The di are the decimal digits of the account number and f(d) is the sum of the decimal digits of 2d (for example, f(7) = 5 because 2 x 7 = 14, and 1 + 4 = 5). For example, 17327 is valid because 1 + 5 + 3 + 4 + 7 = 20, which is a multiple of 10. Implement the function f and write a program to take a 10-digit integer as a command line argument and print a valid 11-digit number with the given integer as its first 10 digits and the checksum as the last digit. Check that when applying the formula to the 11-digit number, the result after mod 10 is 0.
Computers and Technology
1 answer:
Aleksandr [31]1 year ago
8 0

Answer:

Here is the JAVA program:

import java.util.Scanner; //to import Scanner class

public class ISBN

{   public static void main(String[] args)  { // start of main() function body

   Scanner s = new Scanner(System.in); // creates Scanner object

//prompts the user to enter 10 digit integer

   System.out.println("Enter the digits of an ISBN as integer: ");    

   String number = s.next(); // reads the number from the user

   int sum = 0; // stores the sum of the digits

   for (int i = 2; i <= number.length(); i++) {

//loop starts and continues till the end of the number is reached by i

          sum += (i * number.charAt(i - 1) ); }

/*this statement multiplies each digit of the number with i and adds the value of sum to the product result and stores in the sum variable*/

          int remainder = (sum % 11);  // take mod of sum by 11 to get checksum  

   if (remainder == 10)

/*if remainder is equal to 10 adds X at the end of given isbn number as checksum value */

  { System.out.println("The ISBN number is " + number + "X"); }

  else

// displays input number with the checksum value computed

 {System.out.println("The ISBN number is " + number + remainder); }  }  }  

Explanation:

This program takes a 10-digit integer as a command line argument and uses Scanner class to accept input from the user.

The for loop has a variable i that starts from 2 and the loop terminates when the value of i exceeds 10 and this loop multiplies each digit of the input number with the i and this product is added and stored in variable sum. charAt() function is used to return a char value at i-1.

This is done in the following way: suppose d represents each digit:

sum=d1 * 1 + d2 * 2 + d3 * 3 + d4 * 4 + d5 * 5 + d6 * 6 + d7 * 7 + d8 * 8 + d9 * 9

Next the mod operator is used to get the remainder by dividing the value of sum with 11 in order to find the checksum and stores the result in remainder variable.

If the value of remainder is equal to 10 then use X for 10 and the output will be the 10 digits and the 11th digit checksum (last digit) is X.

If the value of remainder is not equal to 10, then it prints a valid 11-digit number with the given integer as its first 10 digits and the checksum computed by sum % 11 as the last digit.  

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Answer:g

   public static int addOddMinusEven(int start, int end){

       int odd =0;

       int even = 0;

       for(int i =start; i<end; i++){

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           }

           else{

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       }

       return odd-even;

   }

}

Explanation:

Using Java programming language:

  • The method addOddMinusEven() is created to accept two parameters of ints start and end
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  • Using a modulos operator we check for even and odds
  • The method then returns odd-even
  • See below a complete method with a call to the method addOddMinusEven()

public class num13 {

   public static void main(String[] args) {

       int start = 2;

       int stop = 10;

       System.out.println(addOddMinusEven(start,stop));

   }

   public static int addOddMinusEven(int start, int end){

       int odd =0;

       int even = 0;

       for(int i =start; i<end; i++){

           if(i%2==0){

               even = even+i;

           }

           else{

               odd = odd+i;

           }

       }

       return odd-even;

   }

}

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