Armstrong Number between Intervals

C Program to print Armstrong Numbers

A number is said to be armstrong number of order n, if  [ abcd… = pow(a,n) + pow(b,n) + pow(c,n) + pow(d,n)+…..   ]. For example,  lets consider 1634 = 1*1*1*1 + 6*6*6*6 + 3*3*3*3 + 4*4*4*4 . Hence, 1634 is an armstrong number. In this section, we will learn how to write a C program to print armstrong number between intervals.

Armstrong number between Intervals

Algorithm to check armstrong number

  • Step 1: Start
  • Step 2: Declare Variable sum, temp, num
  • Step 3: Read num from User
  • Step 4: Initialize Variable sum=0 and temp=num
  • Step 5: Count the number of digits in the num given as n
  • Step 6: While num != 0
                      let m=num%10
                      sum = sum + (m)^n
                      num = num/10
  • Step 7: IF sum==temp
                It is a Armstrong Number
            ELSE
                It is not a  Armstrong Number
  • Step 8: Stop

 

There are two examples given below to print armstromg number between intervals:

Example 1:

Run
#include<stdio.h>
#include<conio.h>
int main ()
{
  int s = 1, e = 500, num, n, arm = 0, i, sum;
  for (i = s; i <= e; i++)
    {
      num = i;
      sum = i;
      while (num != 0)
	{
	  n = num % 10;
	  arm = arm + pow (n, 3);
	  num = num / 10;
	}
      if (sum == arm)
	{
	  printf ("%d\n", i);
	}
      arm = 0;
    }
  return 0;
}

Output:

1
153
370
371
407 

Example 2:

Run
#include<stdio.h>
#include<conio.h>
int main ()
{
  int s = 310, e = 900, num1, n, arm = 0, i, num2, c;
  for (i = s; i <= e; i++)
    {
      num1 = i;
      num2 = i;
      while (num1 != 0)
	{
	  num1 = num1 / 10;
	  ++c;
	}
      while (num2 != 0)
	{
	  n = num2 % 10;
	  arm = arm + (n * n * n);
	  num2 = num2 / 10;
	}
      if (arm == i)
	{
	  printf ("%d\n", i);
	}
      arm = 0;
      c = 0;
    }
  return 0;
}

Output:

370
371
407

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