Sum of digits(updated)

Here is the simple code(in C++) to find the SUM OF DIGITS of the number, until the sum is not in a single digit.

Example:
 
 number = 56

 56==>  5+6=11
 11==>   1+1=2

and 2 is a single digit number.



Best example you can try in this code (Dev-C++) is : 2147483647  and observe the answer.



#include<conio.h>
#include<iostream>
using namespace std;
int main()
{
    int n,nc,rem,sum=0;
    int sum1=10;
    cout<<"enter the number :";
    cin>>n;
    nc=n;

    while(sum1>9)
    {
        sum=0;
        rem=0;
    while(nc>0)
    {
        rem=nc%10;
        sum=sum+rem;
        nc=nc/10;
        sum1=sum;
    }
    nc=sum1;
    cout<<"\n the sum of digits of "<<n<<" is ="<<sum1;
    n=nc;
    }
    getch();
    return 0;
   
}

Output: