Sum of digits of a Number in Java

March 7, 2024

Find the Sum of the Digits of a Number in Java

Given an integer input number, the objective is t0 Find the Sum of the Digits of a Number in Java Language. To do we’ll break the number into it’s digits. Then we add them one by one as we extract them from the number using modulo operator “%”.

Example
Input : 1234
Output : 1 + 2 + 3 + 4 = 10

Find the Sum of the Digits of a Number in Java Language

Given an input integer as the number, our objective is to break down the number into it’s individual digits and sum them up together. To do so we’ll use the follow two operators,

  • Modulo Operator % : We’ll use this to extract the digits from the number.
  • Divide Operator / : We’ll use it to shorten the number by 1 digit.

Once we get the digits, we sum them up one by one with each iteration. Here are some of the methods to solve the above mentioned problem

  • Method 1: Using Brute Force
  • Method 2: Using Recursion I
  • Method 3: Using Recursion II
  • Method 4: Using ASCII Table

We’ll discuss these methods in the upcoming sections below.

Find the Sum of the Digits of a Number in Java

Method 1: Using Brute Force

Jave Code

Run 
public class Main
 {
   public static void main (String[]args)
   {

     int num = 12345, sum = 0;



     //loop to find sum of digits
     while(num!=0){
         sum += num % 10;
         num = num / 10;
     }

     //output
       System.out.println ("Sum of digits : " + sum);
   }


 }

Output

15

Method 2: Using Recursion I

Java Code

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Run 
public class Main
 {
   public static void main (String[]args)
   {

     int num = 12345, sum = 0;
       System.out.println ("Sum of digits : " + getSum (num, sum));
   }

   static int getSum (int num, int sum)
   {

     if (num == 0)
       return sum;

     sum += num % 10;
     return getSum (num / 10, sum);
   }
 }

Output

15

Method 3: Using Recursion II

Java Code

Run 
public class Main
 {
   public static void main (String[]args)
   {

     int num = 12345;
       System.out.println ("Sum of digits : " + getSum (num));
   }

   static int getSum (int num)
   {

     if (num == 0)
       return 0;

     return (num % 10) + getSum (num / 10);
   }

 }

Output

15

Method 4: Using ACSII Table

Java Code

Run 
public class Main
{
  public static void main (String[]args)
  {

    String num = "9876543219876543219876";
    // you can also use BIGINTEGER in place of string
    // if you have to store a number like this
      System.out.println ("Sum of digits : " + getSum (num));
  }

  static int getSum (String num)
  {
    int sum = 0;

    // Traverse through the whole string(char array)
    for (int i = 0; i < num.length (); i++)
      {
	// Ascii value pf numbers start from 48
	// subtracting 48 will give us value in int
	sum = sum + num.charAt (i) - 48;
      }
    return sum;
  }
}

Output

120

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One comment on “Sum of digits of a Number in Java”


  • Pravin

    public class Sum_of_digits_number {
    public static void main(String[] args) {
    String number = “12345”;
    int count = 0;
    for (int i = 0; i < number.length(); i++) {
    count += Integer.parseInt(String.valueOf(number.charAt(i)));
    }
    System.out.println(count);
    }
    }