Armstrong

If the sum of cube of each number is again equal to the number then it is a Armstrong number.

simple logic.

Refer sample Inputs and Outputs.

```
Input 1 : 153
Output: Armstrong Number
Reason((1 ‘ 1 ‘ 1 + 5’5"5 3“3"3=153) which is equal to the number)
Input 2: 134
Output: not a Armstrong
Reason((1 ‘1‘1 + 5‘5’5 2“2"2=134) which is not equal to the number )
```

#include <iostream> using namespace std; int main() { int origNum,num,rem,sum = 0,digit; cout<<"Enter an Integer : "; cin>>origNum; num = origNum; while (num != 0) { digit = num % 10; sum += digit * digit * digit; num /= 10; } if(sum == origNum) cout << "Armstrong Number\n"; else cout<<"Not a Armstrong Number \n"; return 0; }

**Input_1:**

Enter an Integer : 153

**Output:**

Armstrong Number

**Input_2:**

Enter an Integer : 407

**Output:**

Armstrong Number

**Input_3:**

Enter an Integer : 405

**Output:**

Not a Armstrong Number

**Input_4:**

Enter an Integer : 370

**Output:**

Armstrong Number

**Input_5:**

Enter an Integer : 371

**Output:**

Armstrong Number

**Input_6:**

Enter an Integer : 375

**Output:**

Not a Armstrong Number

**Illustration of the Output:**