Let's say, we have a number as N=123456789

If we sum them the result will be 1+2+3+4+5+6+7+8+9 =45 . If we further sum them it will be 4+5 = 9, which is a single digit.

How can we do so? The below is the program that will help us

using System;

namespace ConsoleApplication1

{

class Program

{

static void Main(string[] args)

{

Console.WriteLine(FindSingleDigit(123456789));

Console.ReadKey();

}

public static int FindSingleDigit(int N)

{

var sum = 0;

while (N != 0)

{

sum = sum + (N % 10);

N = N / 10;

}

return sum >= 10 ? FindSingleDigit(sum) : sum;

}

}

}

Let us understand the program first

while (N != 0)

{

sum = sum + (N % 10);

N = N / 10;

}

The loop continues till the value of "N" becomes zero and it gives the total sum of the individual digits. Then it comes out of the loop. Next we perform a checking if the sum obtained is greater than or equals to 10 which is a two digit. In that case , then program makes a recursive call. If it is a single digit, then that value gets return.

Let us analyze the program with an example.

Pass 1: sum = 45

Pass 2: sum = 9