C# Program to Check Disarium Number

Posted in C# Tutorial

Updated on Jan 11, 2024

By Mari Selvan

👁️ 18 - Views

⏳ 4 mins

💬 1 Comment

Photo Credit to CodeToFun

🙋 Introduction

In the realm of programming, exploring mathematical patterns is a fascinating endeavor. One such numeric pattern is the Disarium number.

A Disarium number is a number defined by the sum of its digits each raised to the power of its respective position.

In this tutorial, we'll delve into a C# program designed to check whether a given number is a Disarium number.

📄 Example

Let's explore the C# code that checks whether a given number is a Disarium number.

Program.cs

Copied

using System;

class Program {
  // Function to count the number of digits in a given number
  static int CountDigits(int number) {
    int count = 0;
    while (number != 0) {
      count++;
      number /= 10;
    }
    return count;
  }

  // Function to check if a number is a Disarium number
  static bool IsDisarium(int number) {
    int originalNumber = number;
    int digitCount = CountDigits(number);
    int sum = 0;

    while (number != 0) {
      int digit = number % 10;
      sum += (int) Math.Pow(digit, digitCount);
      digitCount--;
      number /= 10;
    }

    return (sum == originalNumber);
  }

  // Driver program
  static void Main() {
    // Replace this value with your desired number
    int inputNumber = 89;

    // Call the function to check if the number is Disarium
    if (IsDisarium(inputNumber)) {
      Console.WriteLine($"{inputNumber} is a Disarium number.");
    } else {
      Console.WriteLine($"{inputNumber} is not a Disarium number.");
    }
  }
}

💻 Testing the Program

To test the program with different numbers, replace the value of inputNumber in the Main method.

Output

89 is a Disarium number.

Compile and run the program to check if the number is a Disarium number.

🧠 How the Program Works

The program defines a class Program containing static methods CountDigits to count the number of digits and IsDisarium to check if a number is a Disarium number.
The Main method tests the IsDisarium method with a sample number (replace it with your desired number).

📏 Between the Given Range

Let's delve into the C# code that checks for Disarium numbers in the specified range.

Program .cs

Copied

using System;

class Program {
  // Function to check if a number is a Disarium number
  static bool IsDisarium(int number) {
    int originalNumber = number;
    int digitCount = (int) Math.Floor(Math.Log10(number) + 1);
    int sum = 0;

    while (number > 0) {
      int digit = number % 10;
      sum += (int) Math.Pow(digit, digitCount);
      number /= 10;
      digitCount--;
    }

    return sum == originalNumber;
  }

  // Driver program
  static void Main() {
    Console.Write("Disarium numbers in the range 1 to 100: \n");

    for (int i = 1; i <= 100; i++) {
      if (IsDisarium(i)) {
        Console.Write(i + " ");
      }
    }

    Console.WriteLine();
  }
}

💻 Testing the Program

Output

Disarium numbers in the range 1 to 100:
1 2 3 4 5 6 7 8 9 89

Run the program to see the Disarium numbers in the specified range.

🧠 How the Program Works

The program defines a function IsDisarium that checks if a given number is a Disarium number.
Inside the function, it calculates the number of digits in the given number.
It then iterates through each digit, raising it to the power of its position, and calculates the sum.
The main function tests the Disarium condition for numbers in the range from 1 to 100 and prints the results.