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TO GENERATE THE SUM OF ALL DIGITS OF A NUMBER IN C++

Code

#include<iostream>
using namespace std;
int main()
{
int n,i;
int sum=0,num=0;
int rem=0;
cout<<("ENTER THE NUMBER : \n");
cin>>n;
num=n;
while(num>0)
{
	rem=num%10;
	sum=sum+rem;
	num=num/10;
}	
cout<<"THE SUM OF DIGITS OF A NUMBER "<<n<<" IS : "<<sum;
return 0;
}

Output

ENTER THE NUMBER : 543452
THE SUM OF DIGITS OF A NUMBER : 23

What we did - Our Approach

In this problem we have to find the sum of digits of a 'n'^th number.To do so we declare a variable 'sum',which will store the result value(don't forget to initialize it with zero).

Let us consider an example :
Let the number be 23417 so our task is to evaluate 2+3+4+1+7.We can do so by finding the remainder of a number(adding it to sum),divide it with 10 until the number gets 0.For this we need a 'while' loop which will run till 'N>0'.In each iteraion we perform three steps :
1. rem=num%10; //rem=23417%10=7
2. sum=sum+rem; //sum=7
3. num=num/10; //num=23417/10=2341
Hence on repeating these steps(till n>0), we get the sum of digits of a number.