Java Program to Add Digits Until the Number Becomes a Single Digit Number

by Online Tutorials Library July 22, 2022

Java Program to Add Digits Until the Number Becomes a Single Digit Number

In this section, we will create Java programs that add digits of a number until the number becomes a single-digit number. The problem is also known as the digit root problem.

Example

Suppose, 76345 is a number for which we have to find the sum until it becomes a single-digit number.

7 + 6 + 3 + 4 + 5 = 25

Further, we will add 2 and 5.

2 + 5 = 7

We get 7 as a single-digit number.

Brute Force Approach

Calculate the sum of all digits.
Make a recursive call with the sum (calculated in the above step).
If the number is less than 10, return the number.

Let’s implement the above logic in a Java program.

SingleDigit.java

  import java.util.*;  public class SingleDigit   {  //function to calculate the sum of digits       static int digSum(int num)  {  //variable to store sum of digits  int sum = 0;  //loop to do sum while sum is not less than or equal to 9  while (num > 0 || sum > 9)  {  if (num == 0)   {  num = sum;  sum = 0;  }  //determines the last digit of the number and add that digit to the sum variable   sum = sum + num % 10;  //remove the last digit of the number  num = num / 10;  }  //returns the number   return sum;  }  //driver code  public static void main(String args[])  {  int num = 436474;  System.out.println(“The sum of digits is: “+digSum(num));  }  }  

Output:

The sum of digits is: 1

We can also solve the above problem without using any loop and recursion. The approach is as follows:

In this approach, first, check if the number is divisible by 9 or not. If divisible, add the digits together and check if the sum is divisible by the number 9 or not. If divisible, the number is divisible by 9, else not.

For example, 54 i.e. (5 + 4 = 9). We get 9 as the sum of digits. Hence, the sum is divisible by 9.

We observe that if a number (n) is divisible by 9, then sum up its digit until the sum becomes a single digit which is always 9. For example:

n = 2880

Sum of digits = 2 + 8 + 8 + 0 = 18

The number 18 further can be added to form a single digit number.

18 = 1 + 8 = 9

The answer is always 9, if a number has the form like 9x. If the number is of the form 9x+k

A number can be of the form 9x or 9x + k. For the first case, the answer is always 9. For the second case, the answer is always k which is the remainder.

If number is 0, return 0.
Find remainder of number with 9. (number%9).
If remainder is 0, return 9 else return remainder.

  if n == 0     return 0;  if n % 9 == 0            digSum(n) = 9  else                     digSum(n) = n % 9   

Let’s implement the above approach in a Java program.

SumOfDigits.java

  import java.io.*;  public class SumOfDigits   {  static int digSum(int n)  {  if (n == 0)  //if number is equal to 0, return 0  return 0;  return (n % 9 == 0) ? 9 : (n % 9);  }  //driver code  public static void main (String args[])  {  int n = 45678998;  System.out.println(“The sum of digits is: “+digSum(n));  }  }  

Output:

The sum of digits is: 2

The above approach solves the problem in O(1) time. Another approach that solves the problem in O(1) is as follows:

Next TopicNext Greater Number with Same Set of Digits in Java

Java Program to Add Digits Until the Number Becomes a Single Digit Number

Java Program to Add Digits Until the Number Becomes a Single Digit Number

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