Check if a number is an Armstrong number or not in C++

By Raj Gaurav

hello everyone,
In this tutorial, we learn about How to check if a number is an Armstrong number or not, with the help of c++.
before we start, let’s know about what is Armstrong number or how can we say that this is an Armstrong number?
Armstrong number is a number if the sum of the cubes of its digits is equal to the number itself.

let’s see an example:
153 = 1 *1 *1 + 5*5*5 + 3*3*3
=  1 + 125 + 27
=  153
So, 153 is an Armstrong number, because the result is equal to the number itself.
lets see another example also,

371 =  3*3*3 + 7*7*7 + 1*1*1
= 27 + 343 + 1
= 371

How to check if a number is an Armstrong number or not in C++

let’s see the code snippet:

#include<iostream>
using namespace std;
int main ()
{
    int num, temp, rem, sum = 0;
    cout << "Enter a number: ";
    cin >> num;
    temp = num;
    while (temp != 0)
    {
     rem = temp % 10;
     sum = sum + rem*rem*rem;
     temp = temp / 10;
    }
    if (sum == num)
        cout  << " It is an Armstrong number.";
    else
        cout 
     << " It is not an Armstrong number.";
    return 0;
}

in the above piece of code, iostream is used for input /output stream, int main() tells the compiler that function will return an integer value.
A number which is entered by the user is stored in the variable num, we store num into temp for using later in the program,

while (temp != 0)

• it is used to find the number of digits in a given number.

if (sum == num)

• it is used to check whether the sum is equal to the number or not.

rem = temp % 10;

• rem stores every individual digit of that number which is entered by the user

temp = temp / 10;

• Every time temp divided by 10 and the outcome will store in temp if temp becomes zero the loop will be ended.

output: 
Enter a number: 371
It is an Armstrong number.

Do let me know in the comments section if you have any doubts.

Also read:

• How to check whether date is valid or not in C++