Check if a number is special or not in Java

Today we will how to check whether a number is a Special number or not in Java.

But before that, we should know what is a special number.

A special number is a number such that the sum of its digit’s factorial is equal to the number itself.

Sounds confusing?

Let’s take an example to get a clear idea about a special number.

Example: 145

1! + 4! + 5! =145

5!= 5*4*3*2*1=120

4!=4*3*2*1=24

1!=1

145+24+1=145

Now we know what is a special number, so let’s start by learning how to find factorial of a method.

int fact(int n)
    {
        if(n==1 || n==0)
        {
            return 1;
        }
        else
        {
            return n*fact(n-1);
        }
    }

In the above code, we are using the recursion approach, that is, a function calling itself.

If you are not comfortable with recursion you can use the simple way of finding factorial by multiplying the number with its predecessor until 1 comes.

Now let’s look at the complete code:

class test
{
    int fact(int n)
    {
        if(n==1 || n==0)
        {
            return 1;
        }
        else
        {
            return n*fact(n-1);
        }
    }
}
public class MyClass 
{
   public static void main(String args[])
    {
        test t1=new test();
        int n=145;
        int num=n;
        int res=0;
        int rem;
        while(n>0)
        {
            rem=n%10;
            res=res+t1.fact(rem);
            n=n/10;
            
        }
        if(res==num)
        {
            System.out.println("Number is special");
        }
        else
        {
            System.out.println("Number is Not special");
        }
    }
}

The above program when executed will give the following output:

Number is special

Hope this helped you out. We are able to check if a number is special or not with Java programming.

Happy Coding.

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