Question

Check whether a number is a Disarium number using recursion.

(A number is said to be the Disarium number when the sum of its digit raised to the power of their respective positions is equal to the number itself.)

1¹ + 7² + 5³ = 1 + 49 + 125 = 175

				
					ENTER THE NUMBER
175
175 IS A DISARIUM NUMBER

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Code

Java

Python

Java

				
					import java.util.Scanner;
public class disariumNumber
{
    public static int CountDigits(long num,int count)
    {
        if(num==0)
        {
            return count;
        }
        else
        {
            count++;
            return CountDigits(num/10,count);
        }
    }

    public static long CalculateSum(long num,int totalDigits)
    {
        if(num==0)
        {
            return 0;
        }
        else
        {

            return (long)(Math.pow(num%10,totalDigits))+CalculateSum(num/10,totalDigits-1);
        }
    }

    public static void main()
    {

        long num=0,sum=0;
        int FrequencyOfDigits=0;
        Scanner sc=new Scanner(System.in);
        System.out.println("ENTER THE NUMBER");
        num=sc.nextLong();

        FrequencyOfDigits=CountDigits(num,0);
        sum=CalculateSum(num,FrequencyOfDigits);
        if(sum==num)
        {
            System.out.println(num+" IS A DISARIUM NUMBER");
        }
        else
        {
            System.out.println(num+" IS NOT A DISARIUM NUMBER");
        }
    }

}

Python

				
					def countNumberDigits(n,count):
    if(n==0):
        return count
    else:
        count=count+1
        return countNumberDigits(n//10,count)

def calculateSum(n,totalNumberOfDigits):
    if(n==0):
        return 0
    else:
        rem=n%10
        return rem**totalNumberOfDigits + calculateSum(n//10,totalNumberOfDigits-1)

number=int(input("Enter a number:"))
totalDigits=countNumberDigits(number,0)
sum=calculateSum(number,totalDigits)
if(sum==number):
    print(number,"is a Disarium number")
else:
    print(number,"is not a Disarium number")