C program to find all disarium numbers from 1 to 100

C program to find all disarium numbers from 1 to 100:

In this post, we will learn how to write a C program that will print all Disarium numbers from 1 to 100. The program we will write will use a loop to find all disarium numbers in between 1 to 100.

With this post, you will learn how to use different types of loops, how to check if a number is disarium number or not and how to do basic mathematical calculations.

Before we move to the final program, let’s understand what is a disarium number and how to check if a number is disarium number or not.

What is a Disarium number:

A number is called a disarium number if the sum of the digits of the number to the power of its position is equal to the number itself. The position of the digits of a number starts from the leftmost position i.e. the position of the leftmost digit is 1, the second leftmost digit is 2 etc.

Let me show you an example.

Example of Disarium number:

175 is a disarium number. The position of 1 is 1, position of 7 is 2 and 5 is 3.The sum of the digits of the number each raised to its position is:

1^1 + 7^2 + 5^3
= 1 + 49 + 125
= 175

That is the number itself. So, it is a disarium number.

Algorithm to find all disarium numbers from 1 to 100:

The program will use the below algorithm to find all disarium numbers from 1 to 100:

• Run a loop from 1 to 100.
• For each number, check if it is a disarium number or not. If yes, print the number.
• To check if a number is a disarium number or not, create a different function.
  • It will take a number as the parameter and return 1 if the number is a disarium number, else it will return 0.
  • It will create a copy of the given number.
  • Using a loop, it will find the length of the number.
  • Using another loop and another copy of the original number, it will find the sum of digits of the number raised to the position of the digit.
  • If the sum is equal to the number, it will return 1. Else, it will return 0.

C program using while loops:

Below is the complete C program:

#include <stdio.h>
#include <math.h>

int findLength(int n)
{
    int length = 0;
    while (n != 0)
    {
        n /= 10;
        length++;
    }
    return length;
}

int isDisarium(int n)
{
    int length = findLength(n);
    int copyNo = n;
    int lastDigit, sum = 0;

    while (copyNo != 0)
    {
        lastDigit = (int)copyNo % 10;
        sum += pow(lastDigit, length);
        length--;
        copyNo /= 10;
    }

    return sum == n;
}

int main()
{
    int i = 1;

    printf("Disarium numbers within 1 to 100 are: ");

    while (i <= 100)
    {
        if (isDisarium(i))
        {
            printf("%d ", i);
        }
        i++;
    }
    printf("\n");

    return 0;
}

• This program uses only while loops to find all disarium numbers between 1 to 100.
• The main method runs first.
  • Inside this method, it uses a while loop that runs for i = 1 to i = 100.
  • For each value of i, it calls isDisarium method to check if the value is a disarium number or not.
  • If yes, it prints the value of i.
  • After each iteration is completed, it increments the value of i by 1.
• The isDisarium method checks if a number is a disarium number or not. It takes a number as the parameter and returns 1 if it is a disarium number. Else, it returns 0.
  • It calls findLength to find the length of the number.
  • It creates a copy of n and stores it in copyNo.
  • By using a while loop, find the last digit of the number, find the power of that digit raised to its position and add it to the sum variable.
  • Return 1 if the sum is equal to the number. Else, return 0.
• The findLength method finds the length of a number.
  • It uses a while loop to remove the last digit of the number on each step. It increments the value of a length variable which is initialized as 0.
  • The while loop stops if the number become 0.
  • once the loop ends, it returns the calculated length.

Output:

If you run this program, it will print all disarium numbers between 1 to 100.

Disarium numbers within 1 to 100 are: 1 2 3 4 5 6 7 8 9 89 

You can change the limit to any number n to find all disarium numbers from 1 to n.

Method 2: By using a for loop:

We can also use a for loop to write the above program as well. The below program uses for loops:

#include <stdio.h>
#include <math.h>

int findLength(int n)
{
    int length = 0;
    for (; n != 0; n /= 10)
    {
        length++;
    }
    return length;
}

int isDisarium(int n)
{
    int length = findLength(n);
    int lastDigit, sum = 0;

    for (int i = n; i != 0; i = i / 10, length--)
    {
        lastDigit = (int)i % 10;
        sum += pow(lastDigit, length);
    }

    return sum == n;
}

int main()
{
    printf("Disarium numbers within 1 to 100 are: ");

    for (int i = 1; i <= 100; i++)
    {
        if (isDisarium(i))
        {
            printf("%d ", i);
        }
    }
    printf("\n");

    return 0;
}

If you run this program, it will print the same output.

Disarium numbers within 1 to 100 are: 1 2 3 4 5 6 7 8 9 89 

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