C#: Check whether a given number (integer) is Oddish or Evenish
Write a C# Sharp program to check whether a given number (integer) is oddish or evenish.
A number is called "Oddish" if the sum of all of its digits is odd, and a number is called "Evenish" if the sum of all of its digits is even.
Example:
120 -> Oddish
321 -> Evenish
43 -> Oddish
4433 -> Evenish
373 -> Oddish
Sample Solution:
C# Sharp Code:
using System;
using System.Linq;
namespace exercises
{
class Program
{
static void Main(string[] args)
{
int n = 120;
Console.WriteLine("Original number: " + n);
Console.WriteLine("Check the said number is Oddish or Evenish! " + test(n));
n = 321;
Console.WriteLine("\nOriginal number: " + n);
Console.WriteLine("Check the said number is Oddish or Evenish! " + test(n));
n = 43;
Console.WriteLine("\nOriginal number: " + n);
Console.WriteLine("Check the said number is Oddish or Evenish! " + test(n));
n = 4433;
Console.WriteLine("\nOriginal number: " + n);
Console.WriteLine("Check the said number is Oddish or Evenish! " + test(n));
}
public static string test(int n)
{
string str = n.ToString();
int total = 0;
for (int i = 0; i < str.Length; i++)
total += int.Parse(str[i].ToString());
return (total % 2 == 0 ? "Evenish" : "Oddish");
}
}
}
Sample Output:
Original number: 120 Check the said number is Oddish or Evenish! Oddish Original number: 321 Check the said number is Oddish or Evenish! Evenish Original number: 43 Check the said number is Oddish or Evenish! Oddish Original number: 4433 Check the said number is Oddish or Evenish! Evenish
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