Find last non-zero digit in n!

Description

For a given integer n, find the last non-zero digit in thefactorial of n.

n! = n * (n-1) * (n-2) * .... * 3 * 2 * 1

Input format:

The first line should be the value of n.

Output format:

In the first line, print the last non-zero digit of n!.

Sample Input:

5

Sample Output:

2

Sample Input:

10

Sample Output:

8

code:

import java.util.*;

public class Source {
public static void main(String[] args) {
Scanner s = newScanner(System.in);
int n = s.nextInt();
System.out.print(lastNonZeroDigit(n));
}

// Method to find the last digit of n!
static int lastNonZeroDigit(int n) {
// Write your code here
}
}

I have tried many times but did not approach insolutions:

my solution

static int lastNonZeroDigit(int n) {    // Write your code here         if (n<=10)            return n;

if (((n / 10) % 10) % 2 == 0)   return (6 * lastNonZeroDigit(n / 5) * (n % 10) %10);else    return (4 * lastNonZeroDigit(n / 5) * (n % 10) % 10);input: 5

output: 5

** correct me where I am wrong **

Answer rating: 100% (QA)

Answer import javautil class Main static int dig 1 1 2 6 4 2 2 4 2 8 static int lastNon0Digitint n i... View the full answer