Q:

Program to determine whether a given number is a Disarium number

belongs to collection: Number Programs

robaa
Save to bookmark
|
2022-11-08
|
  164 
0

Explanation

In this program, we need to check whether the given number is Disarium or not.

Disarium number

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.

For example, 175 is a Disarium number as follows

11 + 72 + 53 = 1 + 49 + 125 = 175

Some of the other examples of Disarium number are 89, 135, 518 etc.

To find whether given number is Disarium or not, calculate the sum of digits powered with their respective positions. If the sum is equal to the original number then, the given number is Disarium number.

Algorithm

  1. calculateLength() counts the digits present in a number.
    1. Use a while loop to check whether the number is not equal to 0.
    2. Divide the number by 10 and increment the variable length by 1.
    3. Return length.
  2. Define and initialize variable num.
  3. Make a copy of num by storing the value of num in n.
  4. Using while loop calculates remainder rem repeatedly by dividing num with 10.
  5. Calculate the value of rem raised to power its position, i.e. remlen and store the computed value in a variable sum.
  6. Check whether sum is equal to number. If yes, then given number is Disarium number. Else, it is not a Disarium number.

Input:

num = 175  

Output:

11 + 72 + 53 = 1 + 49 + 125 = 175
175 is a disarium number
Php programming, Java programming, C programming, c# programming, Python programming, examples, Php programming examples, Java programming examples, C programming examples, c# programming examples, Python programming examples,
login and add your own answer to this question

All Answers

robaa
2022-11-08
need an explanation for this answer? contact us directly to get an explanation for this answer

Python

#calculateLength() will count the digits present in a number  
def calculateLength(n):  
    length = 0;  
    while(n != 0):  
        length = length + 1;  
        n = n//10;  
    return length;  
   
num = 175;  
rem = sum = 0;  
len = calculateLength(num);  
   
#Makes a copy of the original number num  
n = num;  
   
#Calculates the sum of digits powered with their respective position  
while(num > 0):  
    rem = num%10;  
    sum = sum + int(rem**len);  
    num = num//10;  
    len = len - 1;  
   
#Checks whether the sum is equal to the number itself  
if(sum == n):  
    print(str(n) + " is a disarium number");  
else:  
    print(str(n) + " is not a disarium number");

 

Output:

 175 is a disarium number

 

C

#include <stdio.h>  
   
//calculateLength() will count the digits present in a number  
int calculateLength(int n){  
    int length = 0;  
    while(n != 0){  
        length = length + 1;  
        n = n/10;  
    }  
    return length;  
}  
   
int main()  
{  
    int num = 175, sum = 0, rem = 0, n;  
    int len = calculateLength(num);  
      
    //Makes a copy of the original number num  
    n = num;  
      
    //Calculates the sum of digits powered with their respective position  
    while(num > 0){  
        rem = num%10;  
        sum = sum + (int)pow(rem,len);  
        num = num/10;  
        len--;  
    }  
      
    //Checks whether the sum is equal to the number itself  
    if(sum == n)  
        printf("%d is a disarium number", n);  
    else  
        printf("%d is not a disarium number", n);  
   
    return 0;  
}

 

Output:

175 is a disarium number

 

JAVA

public class DisariumNumber  
{  
    //calculateLength() will count the digits present in a number  
    public static int calculateLength(int n){  
        int length = 0;  
        while(n != 0){  
            length = length + 1;  
            n = n/10;  
        }  
        return length;  
    }  
      
    public static void main(String[] args) {  
        int num = 175, sum = 0, rem = 0, n;  
        int len = calculateLength(num);  
          
        //Makes a copy of the original number num  
        n = num;  
          
        //Calculates the sum of digits powered with their respective position  
        while(num > 0){  
            rem = num%10;  
            sum = sum + (int)Math.pow(rem,len);  
            num = num/10;  
            len--;  
        }  
          
        //Checks whether the sum is equal to the number itself  
        if(sum == n)  
            System.out.println(n + " is a disarium number");  
        else  
            System.out.println(n + " is not a disarium number");  
    }  
}

 

Output:

175 is a disarium number

 

C#

using System;                      
public class DisariumNumber  
{  
    //calculateLength() will count the digits present in a number  
    public static int calculateLength(int n){  
        int length = 0;  
        while(n != 0){  
            length = length + 1;  
            n = n/10;  
        }  
        return length;  
    }  
      
    public static void Main()  
    {  
        int num = 175, sum = 0, rem = 0, n;  
        int len = calculateLength(num);  
          
        //Makes a copy of the original number num  
        n = num;  
          
        //Calculates the sum of digits powered with their respective position  
        while(num > 0){  
            rem = num%10;  
            sum = sum + (int)Math.Pow(rem,len);  
            num = num/10;  
            len--;  
        }  
          
        //Checks whether the sum is equal to the number itself  
        if(sum == n)  
            Console.WriteLine(n + " is a disarium number");  
        else  
            Console.WriteLine(n + " is not a disarium number");  
    }  
}

 

Output:

175 is a disarium number

 

PHP

<!DOCTYPE html>  
<html>  
<body>  
<?php  
//calculateLength() will count the digits present in a number  
function calculateLength($n){  
    $length = 0;  
    while($n != 0){  
        $length = $length + 1;  
        $n = intval($n/10);  
    }  
    return $length;  
}  
      
$num = 175;  
$rem = $sum = 0;  
$len = calculateLength($num);  
      
//Makes a copy of the original number num  
$n = $num;  
   
//Calculates the sum of digits powered with their respective position  
while($num > 0){  
    $rem = $num%10;  
    $sum = $sum + pow($rem,$len);  
    $num = intval($num/10);  
    $len--;  
}  
   
//Checks whether sum is equal to the number itself  
if($sum == $n)  
    print($n . " is a disarium number");  
else  
    print($n . " is not a disarium number");  
?>  
</body>  
</html>

 

Output:

175 is a disarium number

need an explanation for this answer? contact us directly to get an explanation for this answer

total answers (1)

Number Programs
This question belongs to these collections

Similar questions

need a help?

find thousands of online teachers now
Program to determine whether a given number is a h... >>