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C : Display Armstrong Number Between Two Intervals

Tutorial by:Manisha Dubey      Date: 2016-06-17 03:45:33

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Example to find all Armstrong numbers between two integers (entered by the user) using loops and if...else statement.

A positive integer is called an Armstrong number of order n if

abcd... = an + bn + cn + dn + ...

In case of an Armstrong number of 3 digits, the sum of cubes of each digits is equal to the number itself. For example:

153 = 1*1*1 + 5*5*5 + 3*3*3  // 153 is an Armstrong number.

 

 
 This program is build on the concept of how to check whether an integer is an Armstrong number or not.

Example: Armstrong Numbers Between Two Integers

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

int main()
{
    int n1, n2, i, temp1, temp2, remainder, n = 0, result = 0;

    printf("Enter two numbers(intervals): ");
    scanf("%d %d", &n1, &n2);
    printf("Armstrong numbers between %d an %d are: ", n1, n2);

    for(i = n1+1; i<n2; ++i)
    {
        temp2 = i;
        temp1 = i;

        // number of digits calculation
        while (temp1 != 0)
        {
            temp1 /= 10;
            ++n;
        }

        // result contains sum of nth power of its digits
        while (temp2 != 0)
        {
            remainder = temp2%10;
            result += pow(remainder, n);
            temp2 /= 10;
        }

        // checks if number i is equal to the sum of nth power of its digits
        if (result == i) {
            printf("%d ", i);
        }

        // resetting values to check Armstrong number for next iteration
        n = 0;
        result = 0;

    }
    return 0;
}

Output

Enter two numbers(intervals): 999
99999
Armstrong numbers between 999 an 99999 are: 1634 8208 9474 54748 92727 93084

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