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:
