# C Program to check Armstrong Number

• Difficulty Level : Basic
• Last Updated : 22 Nov, 2018

Given a number x, determine whether the given number is Armstrong number or not.

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

```abcd... = pow(a,n) + pow(b,n) + pow(c,n) + pow(d,n) + ....
```

Example:

```Input : 153
Output : Yes
153 is an Armstrong number.
1*1*1 + 5*5*5 + 3*3*3 = 153

Input : 120
Output : No
120 is not a Armstrong number.
1*1*1 + 2*2*2 + 0*0*0 = 9

Input : 1253
Output : No
1253 is not a Armstrong Number
1*1*1*1 + 2*2*2*2 + 5*5*5*5 + 3*3*3*3 = 723

Input : 1634
Output : Yes
1*1*1*1 + 6*6*6*6 + 3*3*3*3 + 4*4*4*4 = 1634
```

## C++

 `// C++ program to determine whether the number is``// Armstrong number or not``#include``using` `namespace` `std;`` ` `/* Function to calculate x raised to the power y */``int` `power(``int` `x, unsigned ``int` `y)``{``    ``if``( y == 0)``        ``return` `1;``    ``if` `(y%2 == 0)``        ``return` `power(x, y/2)*power(x, y/2);``    ``return` `x*power(x, y/2)*power(x, y/2);``}`` ` `/* Function to calculate order of the number */``int` `order(``int` `x)``{``    ``int` `n = 0;``    ``while` `(x)``    ``{``        ``n++;``        ``x = x/10;``    ``}``    ``return` `n;``}`` ` `// Function to check whether the given number is``// Armstrong number or not``bool` `isArmstrong(``int` `x)``{``    ``// Calling order function``    ``int` `n = order(x);``    ``int` `temp = x, sum = 0;``    ``while` `(temp)``    ``{``        ``int` `r = temp%10;``        ``sum += power(r, n);``        ``temp = temp/10;``    ``}`` ` `    ``// If satisfies Armstrong condition``    ``return` `(sum == x);``}`` ` `// Driver Program``int` `main()``{``    ``int` `x = 153;``    ``cout << isArmstrong(x) << endl;``    ``x = 1253;``    ``cout << isArmstrong(x) << endl;``    ``return` `0;``}`

Please refer complete article on Program for Armstrong Numbers for more details!

My Personal Notes arrow_drop_up