Finding sum of digits of a number until sum becomes single digit in C++

In this tutorial, we are going to write a program that sums digits of the given number until it becomes a single digit. Let's see an example.

Input −4543

Output −7

Let's see the steps to solve the problem.

Initialize a number.
Initialize the sum to 0.
Iterate until the sum is less than 9.
- Add each digit of the number to the sum using modulo operator
Print the sum

Example

Let's see the code.

Live Demo

#include <bits/stdc++.h>
using namespace std;
void findTheSingleDigit(int n) {
   int sum = 0;
   while(n > 0 || sum > 9) {
      if(n == 0) {
         n = sum;
         sum = 0;
      }
      sum += n % 10;
      n /= 10;
   }
   cout << sum << endl;
}
int main() {
   int n = 4543;
   findTheSingleDigit(n);
   return 0;
}

Output

you execute the above program, then you will get the following result.

We have another simple method to solve the problem. If the given number is divisible by 9, then the answer is 9. Else the number if n % 9.

Example

Let's see the code.

Live Demo

#include <bits/stdc++.h>
using namespace std;
void findTheSingleDigit(int n) {
   if (n == 0) {
      cout << 0;
   }
   else if (n % 9 == 0) {
      cout << 9 << endl;
   }
   else {
      cout << n % 9 << endl;
   }
}
int main() {
   int n = 4543;
   findTheSingleDigit(n);
   return 0;
}