## Algorithm

1. Initialization:
Initialize variables originalNumber, remainder, result to 0.
Initialize a variable n to store the number of digits.

2. Copy the number:
Copy the original value of number to originalNumber for later comparison.

3. Count the number of digits:
Use a loop to count the number of digits in number and store it in n.

4. Calculate Armstrong sum:
Use a loop to extract each digit from originalNumber.
Raise each digit to the power of n and add the result to result.

5. Check for Armstrong number:
If result is equal to the original number, then it is an Armstrong number.
Otherwise, it is not an Armstrong number.

6. Output:
Display whether the number is an Armstrong number or not.

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

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

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
```

## Code Examples

### #1 Code Example-Programing Check Armstrong Number of three digits

```Code - C Programming```

``````#include <stdio.h>
int main() {
int num, originalNum, remainder, result = 0;
printf("Enter a three-digit integer: ");
scanf("%d", &num);
originalNum = num;

while (originalNum != 0) {
// remainder contains the last digit
remainder = originalNum % 10;

result += remainder * remainder * remainder;

// removing last digit from the orignal number
originalNum /= 10;
}

if (result == num)
printf("%d is an Armstrong number.", num);
else
printf("%d is not an Armstrong number.", num);

return 0;
}
``````
Copy The Code &

Output

cmd
Enter a three-digit integer: 371
371 is an Armstrong number.

### #2 Code Example-Programing Check Armstrong Number of n digits

```Code - C Programming```

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

int main() {
int num, originalNum, remainder, n = 0;
float result = 0.0;

printf("Enter an integer: ");
scanf("%d", &num);

originalNum = num;

// store the number of digits of num in n
for (originalNum = num; originalNum != 0; ++n) {
originalNum /= 10;
}

for (originalNum = num; originalNum != 0; originalNum /= 10) {
remainder = originalNum % 10;

// store the sum of the power of individual digits in result
result += pow(remainder, n);
}

// if num is equal to result, the number is an Armstrong number
if ((int)result == num)
printf("%d is an Armstrong number.", num);
else
printf("%d is not an Armstrong number.", num);
return 0;
}``````
Copy The Code &

Output

cmd
Enter an integer: 1634
1634 is an Armstrong number.