Given a well-formed (non-empty, fully valid) string of digits, let the integer N be the sum of digits. Then, given this integer N, turn it into a string of digits. Repeat this process until you only have one digit left. Simple, clean, and easy: focus on writing this as cleanly as possible in your preferred programming language.

Problem#1

/*Note : Program can take input from binary File*/
#include <iostream>
#include <fstream>
#include <conio.h>
#include <stdio.h>

using namespace std;

int main(){  // jUST fOR fUN Zulqurnain jutt Here
  long int n,m;
	int c=0,a=0;
	printf("Take Input from binary File (y/n) :=:");
	char ch=getch();
	while(1){
		if(ch=='Y'||ch=='y'){
			char r[25];
			cout<<"\n\n\nEnter The Name of the File :=:"; cin>>r;
			printf("\n\n Write New Value or Read Previous ? (W/R) :=:"); 
			char chi=getch();
			if(chi=='W'||chi=='w'){
				int num;
				cout<<"\n\nEnter The Number to store :=: "; cin>>num;
				ofstream of(r,ios::binary|ios::out);
				of.write((char*)&num,sizeof(&num)); // writing
				of.close();
			}
			ifstream in(r,ios::binary|ios::in);
			if(!in){
				cout<<"\n\n\t\t::Wrong Input of file::\n";
			}
			else{
				in.read((char*)&n,sizeof(&n)); // then reading what is written :p
			}
			in.close();
			break;
		}
		else if(ch=='N'||ch=='n'){

			cout<<"\n\n\nEnter Your Number :=:"; cin>>n; cout<<"\n\n";
			break;
		}
		else{
			continue;
		}
	}
	if(n<=9){
		cout<<"Invalid Entry Kindly Number Should be greater than 9\n\n";
		exit(1);
	}
	m = n;
	while(m){
		c+=(m%10);
		m/=10;
	}
	m = c;
	while(m){
		a+=(m%10);
		m/=10;
	}
	cout<<"\n\n\nHere is the Digital Root of Num :"<<n<<": :=:"<<a<<"\n\n";
	return 0;
}