Count the digits of number

Question

package myfirstproject;

import java.util.Scanner;
public class whileloop {

    public static void main(String[] args) {
        // TODO Auto-generated method stub
      long n,sum=0;
      System.out.println("Enter a number:");
      Scanner sc=new Scanner(System.in);
      n=sc.nextLong();
      while(n!=0)
          {
          n/=10;
          sum+=n%10;
      }
      System.out.println("The sum of digits of number is " +sum);
      
      }
    }

output:

enter a number:

For example if I input 745, it adds the first 2 digits and gives the result as 11. But it is not adding the 3rd digit.

Just take input as String, and iterate over each character, get its numeric value and reduce them... — Animesh Sahu, Nov 06 '20 at 08:53
Does this answer your question? [How to sum digits of an integer in java?](https://stackoverflow.com/questions/27096670/how-to-sum-digits-of-an-integer-in-java) — pradeexsu, Nov 06 '20 at 11:07

score 1 · Answer 1 · answered Nov 06 '20 at 08:47

1

Your order is wrong. First n mod ten, then n div 10:

 while(n != 0){
   sum+=n%10;
   n/=10;
 }

It is because you want to get the every single digit of number so you need to modulo it by ten to achieve this.

answered Nov 06 '20 at 08:47

viverte

43
7

score 1 · Answer 2 · answered Nov 06 '20 at 08:51

1

You should interchange the below lines.

  n/=10; 
  sum+=n%10;

The reason is that when you are dividing it by 10, you are losing the last digit before adding it to your sum.

answered Nov 06 '20 at 08:51

Naveen Verma

11
1

score 0 · Answer 3 · answered Nov 06 '20 at 08:51

you need to module and then divide

public static void main(String[] args) {
        long n, sum = 0;
        System.out.println("Enter a number:");
        Scanner sc = new Scanner(System.in);
        n = sc.nextLong();
        while (n != 0) {
            sum+=n%10;
            n/=10;
        }
        System.out.println("The sum of digits of number is " + sum);

    }

Count the digits of number

3 Answers3