How to count even and odd digits in an integer with C#

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// An Automorphic number is a number whose square ends with the same digits 
// as the original number. E.g – 5 : 5 * 5 = 25 //ends with 

using System;

class Program
{
    public static int even = 0, odd = 0;
    
    public static void countEvenOdd(int n) {
        while (n > 0) {
            int reminder = n % 10;
            if (reminder % 2 == 0)
                even++;
            else
                odd++;
            n = n / 10;
        }
    }
    static void Main() {
        int n = 1907834;
      
        countEvenOdd(n);
      
        Console.WriteLine("Total even = " + even);
        Console.WriteLine("Total odd = " + odd);
    }
}




/*
run:
  
Total even = 3
Total odd = 4
  
*/

 



answered Sep 9, 2021 by avibootz

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