Sum of cube of each number is again equal to the number then it is an Armstrong number.

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:

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