Don't be intimidated by factorial to calculate the factorial

Factorial is a very interesting function, the general solution has these:

Package com.threetop.www;
	  Methods for solving factorial public class Jiechen {//Method one: Recursive solution public static int FAC (int n) {int sum; if (n==0| |
	  N==1) {return 1;
	  
	  
	  } SUM=FAC (n-1) *n;
	  
	return sum;
	  }//Method Two: Non-recursive solution (recursive solution) public static int Fac2 (int n) {int sum=1; if (n==0| |
	  N==1) {return 1;
	  for (int i=2;i<=n;i++) {sum*=i;
  return sum;
	  }//Similar to the method of the binary (Gaussian additive method) public static int fac3 (int n) {int begain=1,sum=1; if (n==0| |
	  N==1) {return 1;   
		  } while (true) {if (begain==n)//If the same number is reached, the product of only one number {sum*=n;  else {sum*= (begain*n); Two-way both sides do product operations} if ((N-begain) ==1| |
		  (n-begain) ==0)//termination conditions are either adjacent to two digits or equal to {return sum;       } begain++;
	  Two numbers to the middle close to n--;
	  } public static void Main (String []args) {System.out.println (FAC (10));
	  System.out.println (FAC2 (10));
	  
  System.out.println (FAC3 (10));
 }
}

The results of the operation are as follows:

Two. Question 1: Given an integer n, then n's factorial n. How many 0 are there at the end? For example: N=10,n. =3 628 800,n. Has two 0 at the end.

Question 2: Find N. The position of the lowest bit 1 in the binary representation

The specific implementation is as follows:

Import Java.util.Scanner;  
  
public class Test   
{public  
     static void Main (String args[])  
     {  
      Scanner in = new Scanner (system.in);  
      System.out.print ("Please enter a number that requires factorial:");  
      int N = In.nextint ();  
      int n = n;  
      int sumten = 0;  
      int sumtwo = 0;  
      for (int i=1; i<=n; i++)  
      {  
       int k = i;       
       while (k%5 = = 0)//integer I can be divisible by 5 of the number  
          {  
           Sumten = sumten + 1;  
           K = K/5;  
          }   
       int j = i;  
       while (j%2 = = 0)//integer I can be divisible by 2 of the number  
       {  
        sumtwo = sumtwo + 1;  
        j = j/2;  
       }  
      }  
      The end of SYSTEM.OUT.PRINTLN (n+ "!):" +sumten+ "0");  
      SYSTEM.OUT.PRINTLN (n+ "!) the position of the lowest bit 1 in the binary representation:" +sumtwo);  
     }  
  

The results of the operation are as follows: