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

`abcd... = a`^{n} + b^{n} + c^{n} + d^{n} +

In the case of an Armstrong number of 3 digits, the sum of cubes of each digit is equal to the number itself. For example, 153 is an Armstrong number because

153 = 1*1*1 + 5*5*5 + 3*3*3

Before trying this program, learn how to check whether an integer is an Armstrong number or not.

## Armstrong Numbers Between Two Integers

```
#include <math.h>
#include <stdio.h>
int main() {
int low, high, number, originalNumber, rem, count = 0;
double result = 0.0;
printf("Enter two numbers(intervals): ");
scanf("%d %d", &low, &high);
printf("Armstrong numbers between %d and %d are: ", low, high);
// iterate number from (low + 1) to (high - 1)
// In each iteration, check if number is Armstrong
for (number = low + 1; number < high; ++number) {
originalNumber = number;
// number of digits calculation
while (originalNumber != 0) {
originalNumber /= 10;
++count;
}
originalNumber = number;
// result contains sum of nth power of individual digits
while (originalNumber != 0) {
rem = originalNumber % 10;
result += pow(rem, count);
originalNumber /= 10;
}
// check if number is equal to the sum of nth power of individual digits
if ((int)result == number) {
printf("%d ", number);
}
// resetting the values
count = 0;
result = 0;
}
return 0;
}
```

**Output **

Enter two numbers(intervals): 200 2000 Armstrong numbers between 200 and 2000 are: 370 371 407 1634

In the program, the outer loop is iterated from **(low+ 1)** to **(high - 1)**. In each iteration, it's checked whether `number` is an Armstrong number or not.

Inside the outer loop, the number of digits of an integer is calculated first and stored in `count`

. And, the sum of the power of individual digits is stored in the `result` variable.

If `number` is equal to `result`

, the number is an Armstrong number.

**Note:** You need to reset `count` and `result` to 0 in each iteration of the outer loop.