**Problem Statement:** Given a number N, check if the number is Armstrong’s number or not.

**Examples:**

Example 1:Input:N = 153Output:Armstrong NumberExplanation:153 = 1^3 + 5^3 + 3^3Example 2:Input:N = 154Output:Not Armstrong NumberExplanation:154 ! = 1^3 + 5^3 + 4^3

** Disclaimer**:

*Don’t jump directly to the solution, try it out yourself first.*

## What is Armstrong Number?

A number is thought of as an Armstrong number **if the sum of its own digits raised to the power (number of digits) gives the number itself**.

As you have already seen in the examples above, 153 is an armstrong number because,

- Number of digits in 153 is 3.
- And, sum of every digit raised to the power 3 is 153.

**Approach:**

- First of all, calculate the total number of digits in the given number.
- Once we have the number of digits, run a loop and extract every digit of the number starting from last digit to first digit.
- To extract last digit, divide the number by 10 and the remainder will the required digit. We can simply use to modulo(%) operator to do this.

- Simultaneously, calculate the sum of digits raised to the power (number of digits).
- In the end, check if the sum is the same as original number or not.

**Code:**

## C Program

```
#include<stdio.h>
#include<math.h>
int ArmstrongNumber(int n)
{
int originalno = n;
int count = 0;
int temp = n;
while (temp != 0)
{
count++;
temp = temp / 10;
}
int sumofpower = 0;
while (n != 0)
{
int digit = n % 10;
sumofpower += pow(digit,count);
n /= 10;
}
return (sumofpower == originalno);
}
int main()
{
int n1 = 153;
if (ArmstrongNumber(n1))
{
printf("Yes, it is an Armstrong Number\n");
}
else
{
printf("No, it is not an Armstrong Number\n");
}
return 0;
}
```

**Output:**

Yes, it is an Armstrong Number

**Time Complexity: **O(n) where n is the number of digits since we need to traverse every digit and add digits raised to power no. of digits to sum.

**Space Complexity:** O(1) since no extra space is required

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