Hello geeks,

I'm learning some Java coding and I decided to hit on factorial of numbers.

I'll drop the code and also the factorial for 1000, 2000, ......30000.

1000000! can crash you PC due to the size and time it takes to run.

Let's start with the code:

class Factorial {
 
	public static void main(String[] args) {
		System.out.print("Enter value for factorial: " );
		Scanner input = new Scanner(System.in);
		int value = input.nextInt();
		System.out.println(value + "! is " + factorial(value));
		
		System.out.println("Digit count is " + count(factorial(value)));
	}
	
	static long count(BigInteger k){ //Count all digits
		double factor = Math.log(2)/Math.log(10);
		int count = (int)(factor*k.bitLength()+1);
		if(BigInteger.TEN.pow(count-1).compareTo(k)>0){
			return count-1;
		}
		return count;
	}
	public static BigInteger factorial(long n) {//compute factorial
		 BigInteger result = BigInteger.ONE;
		 for (int i = 1; i <= n; i++)
		 result = result.multiply(new BigInteger(i+"" ));
		 
		return result;
		 }
}

1000

3000

9000!

Enter value for factorial: 9000
9000! is ..........0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
Digit count is 31682


Common 9000 factorial, nairaland complained of post being too long. The zeros are the last digits

11-20000

30000

100000

Good one

Did you use the multiplication method of 1 * 2 * 3... or you used recursion method?

If you use recursive method, it would take a shorter time to run compared to multiplying

Olalekank:
Good one

Did you use the multiplication method of 1 * 2 * 3... or you used recursion method?

If you use recursive method, it would take a shorter time to run compared to multiplying

A program in Java was used.

integer overflow.

Do it recursively

Superstar007:
integer overflow.

yes if you used integers. I used BigInetgers, however.

Would drop the code as well


Edit:

Code added to post