Question

Check whether a number is an Armstrong number using recursion

A positive number is called an Armstrong number if sum of its own digits each raised to the power of the number of digits is equal to number

153
1^3+5^3+3^3=1+125+27=153
153 is an armstrong number

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Code

Java

Python

Java

import java.util.Scanner;
public class ArmstrongNumber
{
    public static int SumOfDigit(int num,int i)
    {
        if(num==0)
        {
            return 0;
        }
        else
        {
            return (int)Math.pow(num%10,i)+SumOfDigit(num/10,i);
        }
    }

    public static void main()
    {
        int num=0,sum=0,noOfDigits=0;
        Scanner sc=new Scanner(System.in);
        System.out.println("ENTER THE NUMBER");
        num=sc.nextInt();
        noOfDigits=Integer.toString(num).length();
        sum=SumOfDigit(num,noOfDigits);
        if(sum==num)
        {
            System.out.println(num+" IS AN ARMSTRONG NUMBER");

        }
        else
        {
            System.out.println(num+" IS NOT AN ARMSTRONG NUMBER");
        }
    }
}

Python

def sumOfDigits(num):
    if(num<=0):
        return 0
    else:
        digit=num%10
        return digit**totalDigits+sumOfDigits(num//10)

#Main Method Starts from here
if(__name__=='__main__'):
    number=int(input("Enter the number:"))
    totalDigits=len(str(number))
    """
    str(number) converted integer to string
    len() gives length of string
    so len(str(number)) combined gives total number of digits
    """
    sum=sumOfDigits(number)
    if(sum==number):
        print(number,"IS AN ARMSTRONG NUMBER")
    else:
        print(number,"IS NOT AN ARMSTRONG NUMBER")